Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 26, 2001, 3–50

THE SIDE-PAIRING ELEMENTS OF MASKIT’S FUNDAMENTAL DOMAIN FOR THE MODULAR GROUP IN GENUS TWO

David Griffiths

King’s College, Department of Mathematics London WC2R 2LS, UK; david.griffiths@entuity.com

Abstract. In this paper we study the hyperbolic geometry on a genus 2 surface. The main object of study is a subset of the set of hyperbolic lengths of closed geodesics on such a surface which arises from an algorithmic choice of shortest loops. Maskit has shown that this data can be used to identify finite sided polyhedral fundamental set for the modular group on the marked hyperbolic surface structures of a given genus. The special nature of genus 2 has made it more accessible than in higher genus and we are able to produce a more detailed picture of the domain and its side-pairing transformations. If the domain can be shown to satisfy certain basic topological criteria, according to a classical theorem of Poincaré, then this would give a set of geometrical generators and relations for the modular group.

0. Introduction

In this paper we study the structural properties of hyperbolic geometry on a genus 2 surface i.e. the crystallographic properties of the Fuchsian groups which uniformise such a surface. Our primary tool, following on from important work of Bernard Maskit ([20]), is a detailed analysis of a type of subset of the set of hyperbolic lengths of closed geodesics on such a surface, which arises from an algorithmic choice of shortest loops in the surface. Maskit shows that this data may be used to identify a finite sided polyhedral fundamental set for the action of the (Teichmüller) modular group on the space of all marked hyperbolic surface structures of a given genus. In genus 2, this action has proved to be more accessible than in higher genus and we are able to produce a more detailed picture of the domain and its side-pairing transformations. If the domain can be shown to satisfy certain basic topological criteria, according to a classical theorem of Poincaré, extended to general discrete group actions, this would then give a set of geometrical generators and relations for the modular group itself.

Maskit’s construction in the special case of genus 2 is as follows. Choose a sequence of 4 non-dividing geodesic loops on the surface satisfying the following intersection property: the second loop intersects the first loop in a single point; the third loop intersects the second loop in a single point, but does not intersect the first; the fourth loop intersects the third loop in a single point, but does not intersect either the first loop or the second loop. We call such an ordered sequence of loops a standard chain. Cutting the surface open along a standard chain we obtain a topological disc and so a standard chain gives a marking for the surface. So our surface, standard chain pair represents a point in Teichmüller space. Now if each choice of geodesic loop was a shortest possible then we say that the standard chain is minimal. We say a surface, standard chain pair lies in the Maskit domain if the standard chain is minimal.

2000 Mathematics Subject Classification: Primary 57M50.

The author was supported by the Swiss National Science Foundation on a Royal Society Exchange Fellowship and by the French Government on a Sejour Scientifique de longue durée.

We wish to consider the intersections of translates of the Maskit domain. Consider an element of the mapping class group. The image of a standard chain under this element is an ordered sequence of loops on the surface. Taking the unique geodesics in the homotopy classes of these loops we obtain another standard chain on the surface. If there exists a surface with both of these standard chains minimal then the Maskit domain and its translate under this mapping class element has non-empty intersection. So solving the problem of which translations have non-empty intersection with the Maskit domain becomes the problem of finding the complete set of allowable minimal standard chain pairs. Due to the special nature of genus 2 surfaces it is known that sequential loops in a standard chain intersect at one of the six Weierstraß points on the surface—the fixed points of the unique hyperelliptic involution that each genus 2 surface exhibits. Theorem 1.1 states that distinct loops in a pair of minimal chains are either disjoint or intersect at Weierstraß points.

Our characterisation of the side-pairing elements of the Maskit domain in genus 2 is as follows: if the Maskit domain has non-empty intersection with a translate under the mapping class group, then this intersection contains a copy of one or other of two special surfaces. One of these special surfaces is the wellknown genus 2 surface with maximal symmetry group. The other special surface does not seem to have appeared in the literature before; it is unusual in that it is not defined by its symmetry group alone, it also requires a certain length equality between geodesic loops to be satisfied. From this characterisation it is a combinatorial exercise to obtain a complete list of mapping class elements that are side-pairing elements of the Maskit domain.

We organise the paper as follows. We begin with general preliminaries concerning genus 2 surfaces and the particular model for Teichmüller space that we adopt. With respect to this model we then repeat Maskit’s definition for a fundamental domain for the Teichmüller modular group. We then construct a one-parameter family of genus 2 surfaces. Two distinguished members of this family are the two special surfaces that feature in our main result. We then show how the main result can be used to give a full list of side-pairing elements of the Maskit domain. We then have the two main technical parts of the paper. In the first we prove the main result under the assumption of Theorem 1.1. In the second we prove Theorem 1.1. We have chosen this order so as to centre the paper on the geometry of the two special surfaces. Moreover we apply results from the first part in the second.

The history of defining a fundamental domain for Mod_{g} for g ≥ 2 goes back to the rough domains of Keen [11]. Maskit covers certain low signature surfaces in his papers [17], [18]. In his doctoral thesis Semmler defined a fundamental domain for closed genus 2 surfaces, based upon locating the shortest dividing geodesic. Recently McCarthy and Papadopoulos [21] have defined a fundamental domain based on the classical Dirichlet construction. For surfaces with one or more punctures there are known triangulations of Teichmüller space. Associated to these are combinatorial fundamental domains—see Harer [14] for an overview of this work. An eventual goal of this work is to give geometrical presentation of the mapping class group in genus 2. The first presentation of the mapping class group in genus 2 was obtained by Birman and Hilden [4] completing the program begun by Bergau and Mennicke [3]. For higher genus surfaces see Hatcher and Thurston [15]. Part of the author’s inspiration for this work came from reading Thurston’s note [25].

The author would like to thank W.J. Harvey, B. Maskit, P. Buser, K.-D. Semmler and C. Bavard for many useful discussions. Further I would like to thank the referee for his or her insight and suggestions.

1. Preliminaries

Throughout our model for the hyperbolic plane H^{2} will be the interior of the unit circle of the complex plane with a metric of constant curvature −1. Likewise 𝒮 will always denote an oriented closed surface of genus 2. The Teichmüller space of genus 2 surfaces 𝒯 is the space of hyperbolic metrics on 𝒮 up to isometries that are isomorphic to the identity. Without further mention, all genus 2 surfaces 𝒮 will be oriented and endowed with a hyperbolic metric.

Let γ denote a simple closed geodesic on 𝒮 . We say that γ is dividing if 𝒮 \γ has two components and non-dividing if 𝒮 \ γ has one component. Throughout the paper ‘\’ denotes ‘set minus’ and by ‘non-dividing geodesic’ we shall always mean ‘simple closed non-dividing geodesic’.

We define a chain to be an ordered set of n non-dividing geodesics 𝒜_{n} = α_{1}, …, α_{n} on 𝒮 such that: |α_{i} ∩ α_{j}| = 1 for |i − j| = 1 and α_{i} ∩ α_{j} = ∅ for |i − j| ≥ 2, where 1 ≤ n ≤ 5 and 1 ≤ i, j ≤ n. A necklace is an ordered set of 6 non-dividing geodesics 𝒜_{6} = α_{1}, …, α_{6} on 𝒮 such that: |α_{i} ∩ α_{j}| = 1 for |i − j| mod 6 = 1 and α_{i} ∩ α_{j} = ∅ for |i − j| mod 6 ≥ 2, where 1 ≤ i, j ≤ 6. We call the geodesics in a chain or necklace the links and we call n, the number of links in a chain, the length of the chain. We note that any length 4 chain extends uniquely to a chain of length 5 and that any chain of length 5 extends uniquely to a necklace, so chains of length 4 and 5 and necklaces can be considered equivalent. We call a chain of length 4 standard and will denote it by 𝒜 .

To a surface, standard chain pair 𝒮 , 𝒜 Maskit associates discrete faithful representation of π_{1}(𝒮) into PSL(2, R); see [20, p. 376]. It is well known that there is a real-analytic diffeomorphism between 𝒟ℱ (𝒮 , PSL(2, R)) and the Teichmüller space 𝒯 (see Abikoff [1]); this diffeomorphism was given explicitly by Maskit in [19]. So there is a one-to-one correspondence between pairs 𝒮 , 𝒜 and points in 𝒯 .

We define a chain 𝒜_{n} = α_{1}, …, α_{n} to be minimal if α_{1} is a shortest nondividing geodesic and if, for any α′_{m} such that 𝒜′_{m} = α_{1}, …, α_{m−1}, α′_{m} is a chain, we have that l(α_{m}) ≤ l(α′_{m}) for 2 ≤ m ≤ n.

Firstly, minimal standard chains exist. To see this we use the fact that given any L > 0 there are only finitely many closed geodesics on 𝒮 that have length ≤ L (see Buser [2, p. 27]). An elementary consequence of this fact is that there are only finitely many shortest non-dividing geodesics; we choose one of them and label it by α_{1} . Choose a non-dividing geodesic that intersects α_{1} exactly once. There are only finitely many shorter non-dividing geodesics with the same intersection property. Choose a shortest and label it by α_{2} . And so on, until we have a minimal standard chain.

Following Maskit we then define 𝒟 ⊂ 𝒯 , the Maskit domain, to be the set of surface, standard chain pairs 𝒮 , 𝒜 with 𝒜 minimal. By the above construction a generic genus 2 surface has exactly one minimal standard chain and so a unique representative on the interior of 𝒟. Maskit also shows that the set of surfaces with more than one minimal standard chain has measure zero in 𝒯 and hence that the boundary of 𝒟 has measure zero. Maskit also gives a proof that the tesselation of 𝒯 by 𝒟 is locally finite. Maskit then observes that 𝒟 satisfies the classical prerequisites to be a fundamental domain for the action of the Teichmüller modular group, or mapping class group, Mod on 𝒯 .

The main question addressed in Maskit’s paper [20] and the author’s paper [13] is the following: given a standard chain, what set of length inequalities must it satisfy in order to be minimal? Maskit, for any genus g, shows that this set is finite and, for genus 2, shows that its cardinality is at most 45. In [13] the author improved this number to 27. The author is confident that this set of inequalities is optimal.

In this paper we examine the tesselation of 𝒯 by 𝒟. More precisely we consider the elements ϕ ∈ Mod that have the property ϕ(𝒟) ∩ 𝒟 ≠ ∅, what we call side-pairing elements of 𝒟. Let ϕ ∈ Mod be a side-pairing element and choose some point 𝒮 ∈ ϕ(𝒟) ∩ 𝒟. So 𝒮 has minimal standard chains 𝒜 , ℬ associated to 𝒟, ϕ(𝒟), respectively. Here ℬ = β_{1}, …, β_{4} where β_{i} = [ϕ(α_{i})], the geodesic in the free homotopy class of ϕ(α_{i}). That is associated to any side-pairing element of 𝒟 there is an ordered pair of minimal standard chains 𝒜 , ℬ on some surface 𝒮 .

Conversely given an ordered pair of minimal standard chains 𝒜 , ℬ on 𝒮 there is an associated side-pairing element of 𝒟. It suffices to calculate a representative ϕ of the unique mapping class such that β_{i} = [ϕ(α_{i})] for i ∈ {1, …, 4}. The natural basis for this calculation is {τ_{i}} for 1 ≤ i ≤ 6 where τ_{i} denotes a left Dehn twist about the link α_{i} in the necklace 𝒜_{6} .

The main fact that enables us to study minimal standard chain pairs is the following: every genus 2 surface 𝒮 exhibits a unique involution, the hyperelliptic involution 𝒥 . This order 2 isometry has six fixed points, the Weierstraß points. Moreover 𝒥 fixes any simple closed geodesic γ on 𝒮 , the action of 𝒥 on γ being classified by the topological type of γ. The restriction of 𝒥 to γ has no fixed points if γ is dividing and two fixed points if γ is non-dividing (see Haas– Susskind [8]). It is a simple consequence that sequential links in a chain intersect at Weierstraß points. We say that two distinct non-dividing geodesics cross if they intersect in a point that is not a Weierstraß point, and we say that two chains cross if a link in one chain crosses a link in the other. We have that:

Theorem 1.1. Minimal standard chains do not cross.

Corollary 1.2. There are only finitely many side-pairing elements.

Proof of Corollary 1.2. Let 𝒜 be a standard chain on 𝒮 . It is enough to show that there are only finitely many other standard chains ℬ on 𝒮 that do not cross 𝒜 . This follows since there are only finitely many non-dividing geodesics that do not cross 𝒜 .

An application of Theorem 1.1 is that τ_{i−1} ∘τ_{i+1} is not a side-pairing element for 1 ≤ i ≤ 4, subscript addition modulo 6. Let 𝒜 be a standard chain and let ℬ = τ_{i−1} ∘ τ_{i+1}(𝒜). Now β_{i} = [τ_{i−1} ∘ τ_{i+1}(α_{i})] crosses α_{i} (see Subsection 2.1 where we perform similar calculations). So, by Theorem 1.1, 𝒜 , ℬ cannot both be minimal.

Given surfaces 𝒮 , 𝒮′ with pairs of minimal standard chains 𝒜 , ℬ and 𝒜′, ℬ′, respectively, we say that 𝒜 , ℬ on 𝒮 is equivalent to 𝒜′, ℬ′ on 𝒮′ if there exists a homeomorphism Ψ: 𝒮 → 𝒮′ such that [Ψ(𝒜)] = 𝒜′, [Ψ(ℬ)] = ℬ′. Our main result in this paper is:

Theorem 1.3. Any minimal standard chain pair is equivalent to a minimal standard chain pair on 𝒪ct or ℰ .

In Subsection 1.2 we construct 𝒪ct and ℰ as members of a one-parameter family of surfaces—each satisfying a certain length equality. Whilst ℰ does not seem to have appeared in the literature before, 𝒪ct is the well-known genus 2 surface of maximal symmetry group.

A simple consequence of Theorem 1.3 is that if φ(𝒟)∩𝒟 ≠ ∅ then ϕ(𝒟)∩𝒟 ∋ 𝒪ct or ℰ . Suppose ϕ(𝒟) ∩ 𝒟 ∋ ∅. Choose a point 𝒮 ∈ ϕ(𝒟) ∩ 𝒟. By the construction above, there exist a minimal standard chain pair 𝒜 , ℬ on 𝒮 such that ℬ = ϕ(𝒜). By Theorem 1.3, 𝒜 , ℬ on 𝒮 is equivalent to 𝒜′, ℬ′ on 𝒪ct or ℰ . It follows that ϕ(𝒟) ∩ 𝒟 ∋ 𝒪ct or ℰ .

The main complaint about the proofs of Theorems 1.3 and 1.1 is that they are based on a case-by-case analysis. That is, we consider cases and derive contradictions using length inequality results for systems of non-dividing geodesics. The majority of the paper is devoted to the proofs of these results. Unfortunately the author has yet to derive a more satisfactory approach.

1.1. Some notation and nomenclature. All of the hyperbolic formulae we use can be found in Buser [2, p. 454]. Given a pair of points X, Y in H^{2} we shall use d(X, Y) to denote the distance between them. For X, Y distinct we shall use ⊥ XY to denote the bisector of X, Y —the set of points Z ∈ H^{2} such that d(Z, X) = d(Z, Y). Given a triplet of distinct points X, Y , Z in H^{2} we shall use ^{}X̸Y Z to denote the angle at the Y vertex of the triangle spanned by X, Y , Z. By a trirectangle we shall mean a compact hyperbolic quadrilateral with three right angles. By a birectangle we shall mean a compact hyperbolic quadrilateral with two adjacent right angles. We shall use curly brackets {∗, ∗, ∗} to indicate unordered sets and round brackets (∗, ∗, ∗) to indicate ordered sets.

1.2. Special surfaces. Suppose we have a trirectangle with acute angle π/4. Label the edges incident upon the π/4 vertex α, β and the edge opposite α (respectively β) by a (respectively b). We label the diagonal from vertex α ∩ β to the vertex a ∩ b by c. Let θ_{a} denote the angle between a, c, et cetera—see Figure 1. We shall abuse notation by using the same symbol as an edge or diagonal to denote its length. We denote such a trirectangle by 𝒬_{α}.

Lemma 1.4. For any given a > cosh^{−1}(\sqrt{2} ) there exists such a trirectangle 𝒬_{α}. Moreover there exist 𝒬_{α} such that c = 2a and c = 2α.

Proof. Firstly a triangle in the hyperbolic plane H^{2} with angles π/4, π/2, 0 has finite edge (between the π/4 vertex and the π/2 vertex) length cosh^{−1} \sqrt{2}.

Consider three geodesics such that the first geodesic intersects the second at an angle π/4 and the second intersects the third at an angle π/2. Let α denote the distance between these intersections. By the above calculation if cosh α = \sqrt{2} the three geodesics bound a π/4, π/2, 0 triangle. So for cosh α > \sqrt{2} there exists a unique common perpendicular between the first and third geodesics. The three geodesics and this common perpendicular now bound a trirectangle.

We now want to show that there exist trirectangles such that c = 2a and c = 2α. By the above we consider the range 2 < cosh^{2} α < ∞. A simple calculation gives

cosh^{2} c − cosh^{2} 2a = −\frac{(cosh^{2} α − 1)(cosh^{4} α − 4 cosh^{2} α + 2)}{cosh^{2} α − 2}.

This expression has exactly one root in the range, cosh^{2} α = 2 + \sqrt{2}. Similarly

cosh^{2} c − cosh^{2} 2α = −\frac{4 cosh^{6} α − 13 cosh^{4} α + 10 cosh^{2} α − 2}{cosh^{2} α − 2}.

Again this expression has exactly one root in the range. Consider the polynomial in the numerator as a polynomial in cosh^{2} α. This polynomial has a root between 2 and 3, and its turning points lie at \frac{1}{2}, \frac{5}{3} . So there exist unique trirectangles such that c = 2a and c = 2α.

We are going to define a fundamental domain in terms of the tesselation of H^{2} by 𝒬_{α}. In Figure 1 we have pictured part of this tesselation, generated by reflecting in each edge. Consider the copy of 𝒬_{α} with its edges and diagonal labelled—i.e. in the negative real, negative imaginary quadrant with its β edge along the real axis. Starting at the a∩b vertex of this trirectangle, in the direction of the a edge: walk a distance 4a; turn right through an angle π − θ_{a}; walk a distance c; turn right through an angle π/2; walk a distance c; and turn right through an angle θ_{a}. Repeat this sequence 3 more times to close the path.

Let Ω_{α} denote the domain circumscribed by this path. Label the sides of Ω_{α} in the order we have walked round them by S_{1} , S′_{3} , S_{4} , S′_{6} , S′_{4} , S′_{2} , S′_{1} , S_{2} , S′_{5} , S_{6} , S_{5} , S_{3} . Define side-pairing elements g_{i} ∈ PSL(2, R) for Ω_{α} so that g_{i}(S_{i}) = S′_{i} for 1 ≤ i ≤ 6. This identification pattern has three length 4 vertex cycles—each with angle sum 2π. It is the same identification pattern as that given by Maskit when constructing a discrete faithful representation to a surface, standard chain pair—see [20, p. 376]. So we obtain a genus 2 surface, with a complete hyperbolic metric, which we shall denote by 𝒮_{α}. We define the octahedral surface 𝒪ct (respectively exceptional surface ℰ ) to be 𝒮_{α} with 𝒬_{α} such that c = 2a (respectively c = 2α).

We need to label a distinguished set of non-dividing geodesics on 𝒮_{α}. Label by ω_{0} , ω_{3} , ω_{0} , ω_{4} , ω_{2} , ω_{1} , ω_{2} , ω_{4} , ω_{0} , ω_{3} , ω_{0} , ω_{4} , ω_{2} , ω_{1} , ω_{2} , ω_{4} the orbits of a ∩ b and α ∩ β on the boundary of Ω_{α} in the order that we walked them and label the origin by ω_{5} . Using the index sets k = 0, 1, 2, 3, l = 4, 5 and modulo 4 addition label by κ_{k,k+1} (respectively κ_{k,l}) the union of orbits of a or b (respectively c) in Ω_{α} passing through ω_{k}, ω_{k+1} (respectively ω_{k}, ω_{l}). Label by λ_{k} the union of orbits of α or β that intersect κ_{k,k+1} . Using the generators g_{i} it is a simple exercise to check that each one of κ_{k,k+1} , κ_{k,l}, λ_{k} projects to a non-dividing geodesic on 𝒮_{α}. Likewise it is easy to check that each set of points ω_{i} projects to a single point on 𝒮_{α}, a Weierstraß point.

Proposition 1.5. The set ⋃ κ_{k,k+1} ∪ κ_{k,l} is the set of shortest non-dividing geodesics on 𝒪ct . The set κ_{1,2} ∪ κ_{3,0} (respectively ⋃ κ_{k,l} ∪ λ_{0} ∪ λ_{2} ) is the set of shortest (respectively second shortest) non-dividing geodesics on ℰ .

It is a simple consequence that minimal chains on 𝒪ct (respectively ℰ ) lie in the set of shortest (respectively shortest and second shortest) non-dividing geodesics.

Proof. Consider ℰ . By definition c = 2α. It follows that 2θ_{b} = θ_{α} and hence α < b. By elementary geometry a < α, b < β and so a < α < b < β.

Take an open disc D_{5} (a circle C_{5} ) of radius c = 2α centred on ω_{5} . No other orbit of a Weierstraß point lies in D_{5} . Around C_{5} , since c = 2α, there are orbits of ω_{k} and of ω_{4} in diametrically opposite pairs. The diameter between the ω_{k} pair projects to κ_{k,5} . The diameters between ω_{4} pairs project to λ_{0} and λ_{2} . So this is the set of shortest non-dividing geodesics passing through ω_{5} . Likewise for ω_{4} .

10 David Griffiths

Figure 1. Construction of the one-parameter family of surfaces.

Now consider an open disc D_{0} (a circle C_{0} ) of radius 2a centred on an orbit of ω_{0} . No other orbit of a Weierstraß point lies in D_{0} and there is a diametrically opposite pair of orbits of ω_{3} on C_{0} . No other orbits of Weierstraß points lie on C_{0} since 2a < c = 2α. So κ_{3,0} , the image of the diameter between the ω_{3} pair, is the shortest non-dividing geodesic passing through ω_{0} . Let C′_{0} denote a circle of radius c = 2α about ω_{0} . There are orbits of ω_{l} in diametrically opposite pairs, projecting to κ_{0,l}. There are no orbits of ω_{1} or ω_{2} on C′_{0} . The nearest such orbit point is at a distance 2b > c = 2α. So κ_{0,l} is the set of second shortest non-dividing geodesics passing through ω_{0} . Likewise for ω_{1} , ω_{2} , ω_{3} .

We now consider 𝒪ct . By definition: c = 2a and so θ_{a} = 2θ_{β}. Suppose that a < b. From the formulae we get that α < β and hence that θ_{a} > θ_{b} and θ_{β} < θ_{α}. It follows that θ_{a} > π/4 and θ_{β} < π/8 giving a contradiction. Likewise for a > b. So a = b.

Take an open disc D_{5} (a circle C_{5} ) of radius c = 2a = 2b centred on ω_{5} . No other orbit of a Weierstraß point lies in D_{5} . Since c = 2a < 2α orbits of ω_{4} lie outside C_{5} . Around C_{5} there are orbits of ω_{k} in diametrically opposite pairs. Again the diameter between the ω_{k} pair projects to κ_{k,5} and so this is the set of shortest non-dividing geodesics passing through ω_{5} . Likewise for ω_{4} .

Now consider an open disc D_{1} (a circle C_{1} ) of radius c = 2a = 2b centred an orbit of ω_{1} . No other orbit of a Weierstraß point lies in D_{1} . Around C_{1} there are orbits of ω_{k} for k = 0, 2 and ω_{l} in diametrically opposite pairs. Again the diameter between the ω_{k} (respectively ω_{l}) pair projects to κ_{0,1} or κ_{1,2} (respectively κ_{1,l}) and so this is the set of shortest non-dividing geodesics passing through ω_{1} . Likewise for ω_{2} , ω_{3} , ω_{0} .

2. Listing of side-pairing elements and proof of Theorem 1.3

In this section show how a listing of side-pairing elements can be generated and prove Theorem 1.3 under the assumption of Theorem 1.1. All minimal chain pairs may be assumed to be non-crossing.

We say that minimal standard chain pair 𝒜 , ℬ is of type (I) (respectively type (II)) if there exist a pair of links Γ (respectively a triplet of links Υ) such that 𝒮 \ Γ (respectively 𝒮 \ Υ) has two components.

The basis of the proof of Theorem 1.3 is to show that a minimal standard chain of type (I) or (II) is equivalent to a standard minimal chain pair on ℰ . To show that a minimal standard chain of neither type (I) nor (II) is equivalent to a standard minimal chain pair on 𝒪ct is a combinatorial exercise.

We label Weierstraß points on 𝒜_{6} so that α_{i} ∋ a_{i}, a_{i+1} . Likewise for ℬ_{6} . Consider a permutation element σ ∈ ℬ_{6} . It is a combinatorial exercise to ennumerate non-equivalent pairs of minimal standard chains on 𝒪ct , ℰ associated to σ. To each of these pairs it is a simple calculation to write down the corresponding side-pairing element of the Maskit domain. We do these exercises for the identity Id and for (i i + 1) which exchanges a_{i}, a_{i+1} for 1 ≤ i ≤ 6.

2.1. Listing of side-pairing elements of the Maskit domain. Let τ_{i} denote a left Dehn twist about α_{i} for 1 ≤ i ≤ 6. It is well known that {τ_{i}} generates the mapping class group—for example Humphries [9] showed that {τ_{i}} for 1 ≤ i ≤ 5 generates it. The action of τ_{i} on 𝒜_{6} is also well known. If j = i or |i − j| > 2 then [τ_{i}(α_{j})] = α_{j}. For j = i − 1 (respectively j = i + 1) and [τ_{i}(α_{j})] is a non-dividing geodesic through a_{j}, a_{i+1} (respectively a_{i}, a_{j+1} ) that does not cross 𝒜_{6} . Moreover α_{j} ∪α_{i}∪[τ_{i}(α_{j})] bounds a pair of triangles that are exchanged under 𝒥 . The geodesics α_{j}, α_{i}, [τ_{i}(α_{j})] lie in anticlockwise order around each triangle. Moreover we know that τ_{i}, τ_{j} commute if |i − j| ≥ 2.

Consider 𝒜 , ℬ on ℰ given by α_{1} = β_{1} = κ_{3,0} , α_{2} = β_{2} = κ_{0,4} , α_{3} = λ_{2} , β_{3} = λ_{0} and α_{4} = β_{4} = κ_{2,5} . It is associated to the identity permutation Id since a_{i} = b_{i} for 1 ≤ i ≤ 6. The corresponding side-pairing element is ι = (τ_{2} ∘ τ_{1})^{3} . We have illustrated the calculation to show that ℬ = ι(𝒜) in Figure 2. The first picture shows τ_{1}(𝒜); the second τ_{2} ∘ τ_{1}(𝒜); et cetera. We now note that Γ = α_{3} ∪ β_{3} is a pair of links that divide 𝒮 into two components, i.e., 𝒜 , ℬ is of type (I).

Figure 2. The action of ι = (τ_{2} ∘ τ_{1})^{3} on the standard chain 𝒜

Now consider 𝒜 , ℬ on ℰ given by α_{1} = β_{1} = κ_{3,0} , α_{2} = κ_{0,4} , β_{2} = κ_{3,4} , α_{3} = β_{3} = λ_{2} and α_{4} = β_{4} = κ_{2,5} . Here 𝒜 , ℬ is associated to (12) since a_{1} = b_{2} , a_{2} = b_{1} and a_{i} = b_{i} for 3 ≤ i ≤ 6. The corresponding side-pairing element is τ_{1} .

Let Υ = α_{2} ∪ β_{1} ∪ β_{2} . We note that 𝒮 \ Υ has three components: two triangles and a torus with boundary component. Also associated to (12) is τ_{1} ∘ ι.

Next consider 𝒜 , ℬ on 𝒪ct given by α_{1} = κ_{3,0} , β_{1} = κ_{0,4} , α_{2} = β_{2} = κ_{3,4} , α_{3} = κ_{2,4} , β_{3} = κ_{2,3} and α_{4} = β_{4} = κ_{1,2} , which is associated to (23). The corresponding side-pairing element is τ_{2} .

Consider 𝒜 , ℬ on ℰ given by α_{1} = β_{1} = κ_{3,0} , α_{2} = κ_{0,5} , β_{2} = κ_{0,4} , α_{3} = β_{3} = λ_{2} , α_{4} = κ_{2,4} and β_{4} = κ_{2,5} , which is associated to (34). The corresponding side-pairing element is τ_{1}^{−2} ∘ τ_{3} . Let Υ = α_{2} ∪ β_{2} ∪ β_{3} . We note that 𝒮 \ Υ has two components: a quadrilateral disc and an annulus. So 𝒜 , ℬ is of type (II). Also associated to (34) is τ_{3} , τ_{3} ∘ τ_{5}^{−2} , τ_{1}^{−2} ∘ τ_{3} ∘ τ_{5}^{−2} and τ_{3} ∘ ι, τ_{1}^{−2} ∘ τ_{3} ∘ ι, τ_{3} ∘ τ_{5}^{−2} ∘ ι, τ_{1}^{−2} ∘ τ_{3} ∘ τ_{5}^{−2} ∘ ι.

Similarly τ_{4} is associated to (45); τ_{5} , ι ∘ τ_{5} are associated to (56); and τ_{6} is associated to (61). The reader can verify that—up to inverses—we have given each side-pairing element of the mapping class group associated to each of the stated permutation elements.

2.2. Projection to the quotient. The quotient of 𝒮 by the hyperelliptic involution 𝒥 is a sphere with six order two cone points 𝒮 /𝒥 . By orbifold we shall always mean a sphere with six order two cone points and a fixed hyperbolic metric. We shall use 𝒪 to denote an orbifold. For technical and pictorial reasons we shall work on the quotient orbifold for the rest of the paper.

The image of a non-dividing geodesic under projection 𝒥: 𝒮 → 𝒪 is a simple geodesic between distinct cone points, what we shall call an arc. Likewise the image of a Weierstraß point under the projection 𝒥: 𝒮 → 𝒪 is a cone point. Definitions of chains, necklaces, links and crossing all pass naturally to the quotient. We define a bracelet Υ to be a set of arcs that contains no crossing arcs, divides 𝒪 and is such that no proper subset of Υ divides 𝒪. As with chains, we call the arcs in a bracelet links and call the number of links the length of a bracelet. In particular a necklace is a bracelet of length 6.

A length 3 bracelet Υ always divides the orbifold into two components, dividing either: one cone point (c) from two; or no cone points from three. For the former we say that Υ cuts off c. For the latter we say that Υ bounds a triangle (the component of 𝒪 \ Υ containing no interior cone points).

On the double cover 𝒮 the lift of Υ divides either: one Weierstraß point c from two; or no Weierstraß points from three. For the former, the single Weierstraß point c lies at the centre of the quadrilateral disc and the two Weierstraß points lie on the interior of the annulus. For the latter, neither triangular disc contains an interior Weierstraß point, whilst the torus with boundary component has three interior Weierstraß points.

We can now restate types (I), (II) on the quotient orbifold. We say that a standard minimal chain pair 𝒜 , ℬ is of type (I) if it contains a length 2 bracelet. We say that a standard minimal chain pair 𝒜 , ℬ is of type (II) if it contains a length 3 bracelet that cuts off a cone point.

Proof of Theorem 1.3. Consider a minimal standard chain pair 𝒜 , ℬ on 𝒪 that is of neither type (I) nor type (II).

We say that an arc set Γ is of type (III) if Γ contains no crossing arcs, each vertex of Γ has index at most four, Γ contains no length 2 bracelets and each length 3 bracelet in Γ bounds a triangle. So the arc set 𝒜 ∪ℬ has property (III).

We say that an arc set Γ on 𝒪 is octahedral if it is graph-isomorphic to a subgraph of the set of shortest arcs on 𝒪ct . We will show that all arc sets of type (III) are octahedral. It follows that 𝒜 , ℬ is equivalent to a standard minimal chain pair on 𝒪ct .

Let Γ be an arc set of type (III). Suppose Γ has a vertex of index four. It is now a simple combinatorial exercise to show that Γ is octahedral. So each vertex of Γ has index at most three. Suppose Γ contains a bracelet of length 3. Again we can show that Γ is octahedral. So each bracelet is of length at least 4. Likewise for Γ containing bracelets of length 4, 5 and 6. So Γ is a tree and we can again show that it is octahedral.

2.3. Arc and cone point labelling and pictorial conventions. In this subsection we define an arc system K ∪ Λ and explain our pictorial conventions. Most length inequality results are given in terms of subsets of this arc system. As its name suggests this arc system is related to the set of non-dividing geodesics we labelled in Subsection 1.2.

Let K be a set of 12 arcs that contains no crossing arcs and has the combinatorial pattern of the edge set of the octahedron. In particular any cone point has four arcs in K incident upon it. Label a pair of cone points having no K arc between them c_{l} for l = 4, 5. We think of c_{4} as being at the South Pole and c_{5} as being at the North Pole. We think of the other cone points as lying on the equator. We label them c_{k} for k = 0, 1, 2, 3 so that there is a K arc between c_{k}, c_{k+1} . Throughout the paper subscript addition for k will be modulo 4. Label the arcs in K so that κ_{k,k+1} is between c_{k}, c_{k+1} and κ_{k,l} is between c_{k}, c_{l}. We define λ_{k} to be the arc between c_{4} , c_{5} that crosses only κ_{k,k+1} ⊂ K. Let Λ = ∪κ_{k}. We now note that the set of non-dividing geodesics we defined on the one-parameter family of surfaces 𝒮_{α} projects to an arc set of the form K ∪ Λ on a one parameter family of orbifolds 𝒪_{α}.

We now explain our pictorial conventions. We always represent the orbifold as a wire-frame figure. Solid (respectively dashed) lines represent arcs in front (respectively behind) the figure. There are three different wire-frames: the octahedral, the exceptional and the triangular prism. The octahedral (respectively exceptional) wire-frame has a wire for each shortest arc (respectively for each shortest and second shortest arc). The triangular prism wire-frame is only used in Section 3. We always represent subsets of K ∪ Λ on the octahedral wire-frame. Any K (respectively Λ) arc in the subset is drawn in thick black (respectively thick grey). We always orient the figure so that c_{4} (respectively c_{5} ) is at the bottom (respectively top). When representing minimal chain pairs, 𝒜 arcs are drawn in thick grey, ℬ arcs are drawn in thick black. We regard α_{i} as oriented from a_{i}, a_{i+1} and use an arrow head to indicate this orientation. Similarly for β_{j}. A single unarrowed thick grey (respectively thick black) line represents the minimal chain 𝒜_{1} = α_{1} (respectively ℬ_{1} = β_{1} ).

We now note that λ_{k} ∪λ_{k+1} is a length 2 bracelet that divides the cone point c_{k} from c_{k+1} , c_{k+2} , c_{k+3} . Likewise Λ_{k} = λ_{k−1} ∪ λ_{k+1} is a length 2 bracelet that divides c_{k}, c_{k+1} from c_{k+2} , c_{k+3} . These arc sets feature in the hypotheses of Lemma 2.3, the result we use to prove Propositions 2.1 and 2.2. Similarly ⋃_{l=4,5} κ_{k,l} ∪λ_{k+1} is a length 3 bracelet that cuts off c_{k+1} . This arc set features in the hypothesis of Theorem 2.6, an important result in the proof Propositions 2.4 and 2.5.

We will denote the two components of 𝒪 \ Λ_{k} by 𝒪_{k,k+1} , 𝒪_{k+2,k+3} so that 𝒪_{k,k+1} ⊃ κ_{k,k+1} . Cutting 𝒪_{k,k+1} open along κ_{k,k+1} we obtain an annulus that we will label by A_{k,k+1} . Let P_{l,k} denote the perpendicular from c_{l} to κ_{k,k+1} in A_{k,k+1} for l = 4, 5. The perpendiculars divide A_{k,k+1} into a pair of birectangles. Denote by 𝒬_{k−1,k} (respectively 𝒬_{k+1,k}) the birectangle such that λ_{k−1} (respectively λ_{k+1} ) lies on its boundary. Similarly for the component 𝒪_{k+2,k+3} .

2.4. Proof of Theorem 1.3 under the assumption of Theorem 1.1

Proposition 2.1. Let 𝒜_{i_{2}}, ℬ_{j_{2}} be a minimal chain pair such that Γ_{i_{2},j_{2}} = α_{i_{2}} ∪ β_{j_{2}} is a length 2 bracelet. Then (i_{2}, j_{2}) = (3, 3) and Γ_{3,3} divides two cone points from two.

Proposition 2.2. Any minimal standard chain pair that contains a length 2 bracelet is equivalent to a minimal standard chain pair on ℰ .

In fact, there is nothing more to prove. To see this, suppose a minimal standard chain 𝒜 , ℬ contains a length 2 bracelet. By Proposition 2.1, Γ_{3,3} = α_{3} ∪β_{3} is this bracelet, Γ_{3,3} divides two cone points from two, and 𝒜 , ℬ contains no other length 2 bracelets. It is now a combinatorial exercise to enumerate standard chain pairs of this kind. Each one of these is equivalent to a minimal standard chain pair on ℰ —see the wire-frames in Figure 3 and two of the wireframes in Figure 12 for some examples.

Figure 3. Minimal chain pairs on ℰ with (i_{2}, j_{2}) = (3, 3).

Lemma 2.3. We have that (i) l(κ_{k,l}) < (l(λ_{k−1}) + l(λ_{k}))/2 for l = 4, 5 and (ii) max{l(κ_{k,k+1}), l(κ_{k+2,k+3})} <(l(λ_{k−1}) + l(λ_{k+1}))/2.

We have pictured the arc sets for Lemma 2.3 with k = 3 in Figure 4.

Proof. (i) One component of 𝒪\λ_{k−1}∪λ_{k} contains c_{k}, label it by 𝒪_{k}. Cut 𝒪_{k} open along κ_{k,l} for l = 4 or 5. The resulting triangular domain has edge lengths 2l(κ_{k,l}), l(λ_{k−1}), l(λ_{k}). By the triangle inequality 2l(κ_{k,l}) < l(λ_{k−1}) + l(λ_{k}).

(ii) Consider the birectangle 𝒬_{k−1,k}. Its κ_{k,k+1} edge is strictly shorter than its λ_{k−1} edge. Likewise for the birectangle 𝒬_{k+1,k}. Adding up edge lengths we have 2l(κ_{k,k+1}) < l(λ_{k−1}) + l(λ_{k+1}). Likewise for the birectangles 𝒬_{k−1,k+2} , 𝒬_{k+1,k+2} .

Proof of Proposition 2.1. Up to relabelling we may suppose that i_{2} ≤ j_{2} . We have that {a_{i_{2}}, a_{i_{2}+1}} = {b_{j_{2}}, b_{j_{2}+1}} and so each one of a_{1}, …, a_{i_{2}−1}, b_{1}, …, b_{j_{2}−1} must lie in one or other component of 𝒪 \ Γ_{i_{2},j_{2}} .

First: Γ_{i_{2},j_{2}} divides two cone points from two. Suppose j_{2} = 4. Each one of b_{1} , b_{2} , b_{3} lies in one or other component of 𝒪 \ Γ_{i_{2},4} . So b_{2} lies in a different component of 𝒪 \Γ_{i_{2},4} to b_{1} or b_{3} and so β_{1} or β_{2} crosses Γ_{i_{2},4} —a contradiction.

We need to derive a contradiction for i_{2} ≤ 2, otherwise 2 < i_{2} ≤ j_{2} < 4 and (i_{2}, j_{2}) = (3, 3). Claim: l(α_{i_{2}}) = l(β_{j_{2}}). If i_{2} = 1, j_{2} ≥ 1 then α_{i_{2}} , β_{j_{2}} are both shortest arcs: by definition if j_{2} = 1 and because ℬ′_{j_{1}} = β_{1}, …, β_{j_{2}−1}, α_{1} is a chain for j_{2} > 1. Similarly, if i_{2} = 2, j_{2} ≥ 2 then both of 𝒜′_{2} = α_{1}, β_{j_{2}} , ℬ′_{j_{1}} = β_{1}, …, β_{j_{2}−1}, α_{2} are chains.

By Lemma 2.3(ii) we have a contradiction if α_{i_{2}}, β_{j_{2}} are both shortest arcs. So i_{2} = 2. The arc α_{1} lies in one component of 𝒪 \ Γ_{2,j_{2}} . Let α′_{2} denote the arc disjoint from Γ_{2,j_{2}} in this component of 𝒪\Γ_{2,j_{2}} . By Lemma 2.3(ii) l(α′_{2}) < l(α_{2}). Since 𝒜′_{2} = α_{1}, α′_{2} is a chain, we have a contradiction.

Next: Γ_{i_{2},j_{2}} divides one cone point c from three. Let 𝒪_{c} denote the component of 𝒪 \ Γ_{i_{2},j_{2}} containing c and let 𝒪′_{c} denote its complement. As above we can show that l(α_{i_{2}}) = l(β_{j_{2}}). Again, if i_{2} = 1 then α_{1}, β_{j_{2}} are both shortest arcs and Lemma 2.3(i) gives a contradiction.

Suppose i_{2} = 2. If c = a_{1} then α_{1} ⊂ 𝒪_{c}. Let α′_{2} be the other arc in 𝒪_{c}. Then by Lemma 2.3(i): l(α′_{2}) < l(α_{2}) and since 𝒜′_{2} = α_{1}, α′_{2} is a chain we have a contradiction. Suppose c ≠ a_{1} , α_{1} ⊂ 𝒪′_{c}. Let α′_{2} be the arc in 𝒪_{c} between a_{2}, c. Again l(α′_{2}) < l(α_{2}), 𝒜′_{2} = α_{1}, α′_{2} is a chain and we have a contradiction.

Finally, consider i_{2} > 2. Each of α_{1}, …, α_{i_{2}−1} must lie in 𝒪′_{c}, otherwise one of these arcs would cross Γ_{i_{2},j_{2}} . Let α′_{i_{2}} be the arc in 𝒪_{c} between a_{i_{2}}, c. Again l(α′_{i_{2}}) < l(α_{i_{2}}), 𝒜′_{i_{2}} = α_{1}, …, α_{i_{2}−1}, α′_{i_{2}} is a chain and we have a contradiction.

Proposition 2.4. If Υ_{i_{3},j_{3}} = α_{i_{3}} ∪ β_{j_{3}−1} ∪ β_{j_{3}} is a length 3 bracelet that cuts off a cone point c, then we have that i_{3} > 1, j_{3} > 2; if (i_{3}, j_{3}) = (2, 3), then (a_{2}, a_{3}) = (b_{2}, b_{4}), c = a_{1} = b_{1} ; if (i_{3}, j_{3}) = (2, 4), then (a_{2}, a_{3}) = (b_{5}, b_{3}), a_{1} = c ∉ {b_{1}, b_{2}}; if (i_{3}, j_{3}) = (3, 3), then (a_{3}, a_{4}) = (b_{4}, b_{2}), b_{1} = c ∉ {a_{1}, a_{2}}; if (i_{3}, j_{3}) = (3, 4), then (a_{3}, a_{4}) = (b_{3}, b_{5}), c ∉ {a_{1}, a_{2}} = {b_{1}, b_{2}}; if (i_{3}, j_{3}) = (4, 3), then (a_{4}, a_{5}) = (b_{4}, b_{2}), b_{1} = c ∉ {a_{1}, a_{2}}; and if (i_{3}, j_{3}) = (4, 4), then (a_{4}, a_{5}) = (b_{3}, b_{5}), c ∉ {a_{1}, a_{2}} = {b_{1}, b_{2}}.

Proposition 2.5. Any minimal standard chain pair that contains a length 3 bracelet that cuts off a cone point is equivalent to a minimal standard chain pair on ℰ .

Unlike Proposition 2.2 which followed directly from Proposition 2.1, Proposition 2.5 does not follow directly from Proposition 2.4; there is still something to prove. However almost all the arguments reproduce arguments given in the proof of Proposition 2.4. The main result we apply to prove Proposition 2.4 is Theorem 2.6 which appeared in paper [12] as Theorem 1.1. Theorem 2.7 also appeared in paper [12] as Theorem 1.2.

Theorem 2.6. Suppose, for some k, that l(κ_{k,l}) ≤ l(κ_{k+1,l}), l(λ_{k+1}) ≤ l(λ_{k−1}) for l = 4, 5, then l(κ_{k,l}) = l(κ_{k+1,l}), l(λ_{k+1}) = l(λ_{k−1}) for l = 4, 5.

Theorem 2.7. Suppose, for some k, that κ_{k,l} is a shortest arc for l = 4, 5 and that l(κ_{k,k+1}) ≤ l(κ_{k+2,k+3}), l(λ_{k+1}) ≤ l(λ_{k−1}). Then 𝒪 is the octahedral orbifold.

Figure 4. Arc sets for Lemma 2.3 and Theorems 2.6, 2.7 with k = 3

Lemma 2.8. Suppose, for some k, that l(κ_{k,l}) = l(κ_{k+1,l}), l(λ_{k+1}) = l(λ_{k−1}) for l = 4, 5. Then l(κ_{k,4}) = l(κ_{k+1,5}) if and only if l(κ_{k+2,4}) = l(κ_{k+3,5}).

Proof. We will show l(κ_{k+2,4}) = l(κ_{k+3,5}) implies l(κ_{k,4}) = l(κ_{k+1,5}). The other direction follows similarly.

Since l(λ_{k+1}) = l(λ_{k−1}) the annulus 𝒜_{k+2,k+3} has mirror symmetry exchanging the birectangles 𝒬_{k−1,k+2} , 𝒬_{k+1,k+2} . That is P_{4,k+2} , P_{5,k+2} are equally spaced about the geodesic boundary component κ_{k+2,k+3} of 𝒜_{k+2,k+3} . We know that c_{k+2} , c_{k+3} are also equally spaced about this boundary component. Since l(κ_{k+2,4}) = l(κ_{k+3,5}) it follows that l(P_{4,k+2}) = l(P_{5,k+2}). That is 𝒜_{k+2,k+3} has rotational symmetry exchanging 𝒬_{k−1,k+2} , 𝒬_{k+1,k+2} . Gluing along κ_{k+2,k+3} to recover 𝒪_{k+2,k+3} this symmetry is respected. This in turn implies that 𝒪_{k,k+1} has rotational symmetry—c.f. the proof of Theorem 1.2 in [12]—and hence that l(κ_{k,4}) = l(κ_{k+1,5}).

Proof of Proposition 2.4. Suppose Υ is a length 3 bracelet that cuts off a cone point c. Label arcs in Υ by κ_{k,l}, λ_{k+1} for l = 4, 5. This labelling then extends uniquely to K_{k}∪Λ_{k} ⊂ K∪Λ where K_{k} = ⋃_{l=4,5} κ_{k,l}∪κ_{k+1,l}∪κ_{k,k+1}∪κ_{k+2,k+3} . To see this we proceed as follows. Label by 𝒪_{c} the component of 𝒪 \ Υ containing c. Set c = c_{k+1} and then label the arcs in 𝒪_{c} between c_{k+1} , c_{l} by κ_{k+1,l} for l = 4, 5 and between c_{k}, c_{k+1} by κ_{k,k+1} . Let 𝒪′_{c} denote the component of 𝒪 \ Υ not containing c. Label by λ_{k−1} the arc in 𝒪′_{c} between c_{4} , c_{5} such that ⋃_{l=4,5} κ_{k,l} ∪ λ_{k−1} bounds a triangle. Label by κ_{k+2,k+3} the arc disjoint from Υ in 𝒪′_{c}. We will use this extension of arc labelling for applications of Theorem 2.6.

Suppose i_{3} = 1. In Figure 5 the four wire-frames represent all the configurations of 𝒜_{1}, ℬ_{j_{3}} such that Υ_{1,j_{3}} cuts off a cone point. For all but the third configuration we can use Theorem 2.6 and Lemma 2.3(ii) to derive a contradiction. For the third configuration we can apply Theorem 2.7 to show that 𝒪 is the octahedral orbifold. As we have observed before minimal chains on 𝒪ct lie in its set of shortest arcs. Any length 3 bracelet in this set bounds a triangle.

Consider, for example the fourth configuration, with j_{3} = 4. Set α_{1} = κ_{3,4} , β_{3} = λ_{0} , β_{4} = κ_{3,5} . This extends uniquely to K_{3} ∪ Λ_{3} . We note that ℬ′_{3} = β_{1}, β_{2}, α_{1} , ℬ′_{4} = β_{1}, β_{2}, β_{3}, κ_{0,5} are both chains and so λ_{0} is a shortest arc and l(κ_{3,5}) ≤ l(κ_{0,5}). We know that κ_{3,4} is a shortest arc, so the hypotheses of Theorem 2.6 are satisfied. So l(λ_{0}) = l(λ_{2}). By Lemma 2.3(ii): l(κ_{3,0}) < l(λ_{0}) = l(λ_{2}) which contradicts λ_{0} being a shortest arc.

For the third configuration we argue as follows. Set α_{1} = κ_{3,4} , β_{2} = κ_{3,5} , β_{3} = λ_{0} which extends uniquely to K_{3} ∪ Λ_{3} . We note that ℬ′_{2} = β_{1}, α_{1} , ℬ′_{3} = β_{1}, β_{2}, λ_{2} are chains, so κ_{3,5} is a shortest arc and l(λ_{0}) ≤ l(λ_{2}). So the hypotheses of Theorem 2.7 are satisfied: 𝒪 is the octahedral orbifold.

Figure 5. Configurations of 𝒜_{1}, ℬ_{j_{3}}

So we have shown that i_{3} > 1. Next we show that j_{3} > 2 (Figure 6). We then consider j_{3} = 3 (Figure 7), then j_{3} = 4 (Figure 8). By cone point labels on Υ_{i_{3},j_{3}} we know that {a_{i_{3}}, a_{i_{3}+1}} = {b_{j_{3}−1}, b_{j_{3}+1}}. Suppose (a_{i_{3}}, a_{i_{3}+1}) = (b_{j_{3}+1}, b_{j_{3}−1}). If a_{i_{3}−1} = b_{j_{3}} then α_{i_{3}−1}, β_{j_{3}} share endpoints. Unless (i_{3}, j_{3}) = (4, 3), by Proposition 2.1, α_{i_{3}−1} = β_{j_{3}} . So Υ_{i_{3},j_{3}} = β_{j_{3}−1} ∪ α_{i_{3}−1} ∪ α_{i_{3}} with (b_{j_{3}−1}, b_{j_{3}}) = (a_{i_{3}+1}, a_{i_{3}−1}) which is covered by the argument we give for (i′_{3}, j′_{3}) = (j_{3} − 1, i_{3}) since b_{j_{3}−2} ≠ a_{i_{3}} = b_{j_{3}+1} . Suppose (i_{3}, j_{3}) = (4, 3), (a_{4}, a_{5}) = (b_{5}, b_{3}) and a_{3} = b_{4} . Suppose α_{3} = β_{3} . If b_{1} = c then 𝒜 , ℬ_{3} is equivalent to a minimal chain pair on ℰ ; see the third wire-frame, Figure 12. Otherwise, it is covered by an argument we give for (i_{3}, j_{3}) = (2, 4) since b_{1} ≠ a_{4} = b_{5} . If α_{3}∪β_{3} is a bracelet then 𝒜 , ℬ_{3} is equivalent to a minimal chain pair on ℰ by Proposition 2.2; see the fourth wire-frame, in Figure 12. So, for (a_{i_{3}}, a_{i_{3}+1}) = (b_{j_{3}+1}, b_{j_{3}−1}), we may suppose that a_{i_{3}−1} ≠ b_{j_{3}} .

For j_{3} = 2 we begin by i_{3} = 2. First we consider (a_{2}, a_{3}) = (b_{1}, b_{3}). If a_{1} ≠ b_{2} then 𝒜′_{2} = α_{1}, β_{1} is a chain and α_{2} is a shortest arc—this is equivalent to i_{3} = 1. So a_{1} = b_{2} , α_{1}, β_{1} share endpoints and so, by Proposition 2.1, α_{1} = β_{1} ; see the first wire-frame, Figure 6. We can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of α_{1} = β_{1} . Next (a_{2}, a_{3}) = (b_{3}, b_{1} ). We may suppose that a_{1} ≠ b_{2} . So a_{1} = c or a_{1} ∈ 𝒪′_{c}; see the second and third wireframes. We can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{1}, β_{2} , respectively. For the latter , with labelling so that β_{2} = λ_{0} , we show that l(κ_{3,0}) < l(λ_{0}) = l(λ_{2}) which gives a contradiction since ℬ′_{2} = β_{1}, κ_{3,0} is a chain.

Next i_{3} = 3. Suppose (a_{3}, a_{4}) = (b_{1}, b_{3}). If {a_{1}, a_{2}} ∌ b_{2} then 𝒜′_{3} = α_{1}, α_{2}, β_{1} is a chain and α_{3} is a shortest arc—again equivalent to i_{3} = 1. So {a_{1}, a_{2}} ∋ b_{2} . If a_{1} = b_{2} then {a_{1}, a_{3}} = {b_{1}, b_{2}} and so β_{1} ∪α_{1} ∪α_{2} is a bracelet. Since 𝒜′_{2} = α_{1}, β_{1} is a chain it follows that this is a bracelet of shortest arcs and we can apply Proposition 2.10. So a_{2} = b_{2} . Either a_{1} = c or a_{1} ∈ 𝒪′_{c}; see the fourth and fifth wire-frames. For the former we can apply Theorem 2.7 to show that 𝒪 is the octahedral orbifold. For the latter we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{1} .

Figure 6. Configurations of 𝒜_{i_{3}}, ℬ_{2}

Finally i_{3} = 4. If a_{2} = b_{2} then the arcs α_{1}, β_{1}, α_{2}, β_{2} all have exactly one cone point in common, 𝒜′_{2} = α_{1}, β_{1} , ℬ′_{2} = β_{1}, α_{1} are both chains and so each arc in this set is shortest. We can now apply Lemma 2.11: 𝒪 is the octahedral orbifold. So a_{2} ≠ b_{2} . Suppose (a_{4}, a_{5}) = (b_{1}, b_{3}). Either a_{1} = b_{2} or a_{3} = b_{2} ; see the sixth and seventh wire-frames. We can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{1} . For (a_{4}, a_{5}) = (b_{3}, b_{1} ) we may suppose that a_{3} ≠ b_{2} . So a_{1} = b_{2} and we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{2} ; see the eighth wire-frame.

For j_{3} = 3 we begin with i_{3} = 2. Suppose (a_{2}, a_{3}) = (b_{2}, b_{4}). If a_{1} ≠ b_{1} then 𝒜′_{2} = α_{1}, β_{1} is a chain and α_{2} is a shortest arc, again equivalent to i_{3} = 1. So a_{1} = b_{1} . For a_{1} = b_{1} = c: 𝒜_{2}, ℬ_{3} is equivalent to a minimal chain pair on ℰ ; see the first wire-frame, Figure 12. For a_{1} = b_{1} ∈ 𝒪′_{c} we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{2} ; see the first wire-frame, Figure 7. Now suppose (a_{2}, a_{3}) = (b_{4}, b_{2} ). We may suppose that a_{1} ≠ b_{3} . If a_{1} = b_{1} then {a_{1}, a_{3}} = {b_{1}, b_{2}} and so β_{1} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs. So a_{1} ≠ b_{1} . For a_{1} = c, b_{1} ∈ 𝒪′_{c} we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{2} ; see the second wire-frame.

Suppose a_{1} ∈ 𝒪′_{c}, b_{1} = c; see the third wire-frame. Set α_{2} = κ_{3,4} , β_{2} = κ_{3,5} , β_{3} = λ_{0} , α_{1} = κ_{1,4} which extends uniquely to K_{3} ∪ Λ_{3} ∪ κ_{1,4} ∪ κ_{2,5} . The hypotheses of Theorem 2.6 are satisfied: each one of 𝒜′_{2} = α_{1}, κ_{0,4} , ℬ′_{2} = β_{1}, κ_{0,5} , ℬ′_{3} = β_{1}, β_{2}, λ_{2} is a chain. So l(κ_{3,l}) = l(κ_{0,l}) for l = 4, 5 and l(λ_{0}) = l(λ_{2}). By Lemma 2.3(ii), β_{3} = λ_{0} is not a shortest arc. Below we show that κ_{2,5} is a shortest arc, which gives a contradiction since ℬ′′_{3} = β_{1}, β_{2}, κ_{2,5} is a chain. First ℬ′′_{2} = β_{1}, α_{2} is a chain and so l(β_{2}) ≤ l(α_{2}). Next, κ_{3,0} = β_{1} , κ_{1,4} = α_{1} are both shortest arcs and l(κ_{3,5}) = l(κ_{0,5}). By Lemma 2.9 there exists an arc γ such that l(γ) ≤ l(κ_{3,5}) = l(κ_{0,5}). The arc γ is such that 𝒜′′_{2} = α_{1}, γ is a chain and so l(α_{2}) ≤ l(γ). So, we have that l(κ_{3,5}) = l(β_{2}) ≤ l(α_{2}) = l(κ_{3,4}) ≤ l(γ) ≤ l(κ_{3,5}) and hence l(κ_{3,4}) = l(κ_{3,5}). So l(κ_{3,4}) = l(κ_{0,5}) and by Lemma 2.8, l(κ_{1,4}) = l(κ_{2,5}). Since α_{1} = κ_{1,4} is a shortest arc we are done.

So a_{1} ≠ b_{1} ∈ 𝒪′_{c}; see the fourth wire-frame. Set α_{2} = κ_{3,4} , β_{2} = κ_{3,5} , β_{3} = λ_{0} , β_{1} = κ_{2,3} which extends uniquely to K ∪Λ. We have that 𝒜′_{2} = α_{1}, κ_{0,4} is a chain and so l(κ_{3,4}) ≤ l(κ_{0,4}). Also each one of ℬ′_{2} = β_{1}, α_{2} , 𝒜′′_{2} = α_{1}, β_{3} , ℬ′_{3} = β_{1}, β_{2}, κ_{0,5} is a chain and so l(κ_{3,5}) = l(β_{2}) ≤ l(α_{2}) ≤ l(β_{3}) ≤ l(κ_{0,5}). By Theorem 2.6, l(λ_{2}) ≤ l(λ_{0}). Now both ℬ′′_{3} = β_{1}, β_{2}, λ_{1} , ℬ′′′_{3} = β_{1}, β_{2}, λ_{3} are chains and so l(λ_{0}) ≤ min{l(λ_{1}), l(λ_{3})}. By Theorem 2.12, 𝒪 is the octahedral orbifold.

Next i_{3} = 3. Suppose (a_{3}, a_{4}) = (b_{2}, b_{4}). If {a_{0}, a_{1}} ∌ b_{1} then 𝒜′_{3} = α_{1}, α_{2}, β_{1} is a chain, α_{2} is a shortest arc. So {a_{1}, a_{2}} ∋ b_{1} . If {a_{1}, a_{2}} ∋ b_{3} then α_{1} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. So {a_{1}, a_{2}} ∌ b_{3} . Since α_{1} does not cross Υ_{3,3} it follows that a_{1}, a_{2} lie in the same component of 𝒪 \ Υ_{3,3} and hence a_{1}, a_{2} ∈ 𝒪′_{c}. If a_{1} = b_{1} then β_{1} ∪α_{1} ∪α_{2} is a bracelet of shortest arcs. So a_{2} = b_{1} ; see the fifth wire-frame; and we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{2} . Now (a_{3}, a_{4}) = (b_{4}, b_{2}). Again we may suppose that a_{2} ≠ b_{3} . Suppose a_{1} = b_{3} . If a_{2} = b_{1} then we have that α_{1} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. If a_{2} ≠ b_{1} then ℬ′_{3} = β_{1}, β_{2}, α_{1} , 𝒜′_{2} = α_{1}, β_{3} are both chains and so β_{3}, α_{2} are both shortest arcs. Since (a_{1}, a_{3}) = (b_{3}, b_{4}) it follows that β_{3} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs. So {a_{1}, a_{2}} ∌ b_{3} . Again this implies that a_{1}, a_{2} ∈ 𝒪′_{c}. If b_{1} = c then 𝒜_{3}, ℬ_{3} is equivalent to a minimal chain pair on ℰ ; see the second wire-frame, Figure 12. If b_{1} ∈ 𝒪′_{c} then a_{1} = b_{1} or a_{2} = b_{1} and we can apply Theorems 2.6 and 2.12 to show that 𝒪 is 𝒪ct ; see the sixth and seventh wire-frames.

Next i_{3} = 4. First (a_{4}, a_{5}) = (b_{2}, b_{4}). Suppose {a_{1}, a_{2}, a_{3}} ∌ b_{1} then 𝒜′_{4} = α_{1}, α_{2}, α_{3}, β_{1} is a chain and α_{4} is a shortest arc. So {a_{1}, a_{2}, a_{3}} ∋ b_{1} . Also {a_{1}, a_{2}, a_{3}} ∋ b_{3} since neither α_{1} nor α_{2} crosses Υ_{4,3} . So {a_{1}, a_{2}, a_{3}} ⊃ {b_{1}, b_{3}}. If {a_{1}, a_{2}} = {b_{1}, b_{3}} then α_{1} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. If {a_{2}, a_{3}} = {b_{1}, b_{3}} then α_{2} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. For (a_{2}, a_{3}) = (b_{1}, b_{3}) we have that 𝒜′_{2} = α_{1}, β_{1} , ℬ′_{2} = β_{1}, α_{1} are both chains. For (a_{2}, a_{3}) = (b_{3}, b_{1}) we have that ℬ′_{3} = β_{1}, β_{2}, α_{1} , 𝒜′_{2} = α_{1}, β_{3} , ℬ′_{2} = β_{1}, α_{2} are all chains. So {a_{1}, a_{3}} = {b_{1}, b_{3}} and we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict the shortness of β_{2} ; see the eighth and ninth wire-frames.

Figure 7. Configurations of 𝒜_{i_{3}}, ℬ_{3}

Next (a_{4}, a_{5}) = (b_{4}, b_{2}). We may suppose that a_{3} ≠ b_{3} . Again neither α_{1} nor α_{2} cross Υ_{4,3} so {a_{1}, a_{2}} ∋ b_{3} . If {a_{1}, a_{2}} ∋ b_{1} then α_{1} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. So {a_{1}, a_{2}} ∌ b_{1} . Suppose a_{3} = b_{1} ; see the tenth, eleventh and twelfth wire-frames, Figure 7. In each case ℬ′_{3} = β_{1}, β_{2}, α_{1} , 𝒜′_{2} = α_{1}, β_{3} , ℬ′_{2} = β_{1}, α_{2} are chains and so β_{2} is a shortest arc. We can now apply Theorems 2.6 and 2.12 to show that 𝒪 is 𝒪ct . So {a_{1}, a_{2}, a_{3}} ∌ b_{1} and hence ℬ′_{3} = β_{1}, β_{2}, α_{1} is a chain, β_{3} is a shortest arc. If b_{1} = c; see the 13th and 14th wire-frames. We can apply Theorem 2.6 and Lemma 2.3(ii) to contradict β_{3} being a shortest arc. If a_{1} = c or a_{3} = c (see the 15th and 16th wire-frames) we can again apply Theorems 2.6 and 2.12. The argument differs slightly from that given above. Above we had that l(λ_{2}) ≤ l(λ_{0}) and l(λ_{0}) ≤ min{l(λ_{1}), l(λ_{3})}. Again we have that l(λ_{2}) ≤ l(λ_{0}) but only have l(λ_{0}) ≤ l(λ_{1}). Here, however, we have that κ_{3,0} is a shortest arc. The argument runs as follows; c.f. the proof of Lemma 2.11. Suppose a_{1} = c. Set α_{4} = λ_{0} , β_{2} = κ_{3,5} , β_{3} = κ_{3,4} , α_{2} = κ_{2,3} which extends uniquely to K∪Λ. Since ℬ′_{3} = β_{1}, β_{2}, α_{1} , 𝒜′_{2} = α_{1}, β_{3} are both chains, β_{3} = κ_{3,4} , α_{2} = κ_{2,3} are both shortest arcs. Since ℬ′_{2} = β_{1}, κ_{0,5} is also a chain we can apply Theorem 2.6: l(λ_{2}) ≤ l(λ_{0}). Now 𝒜′_{4} = α_{1}, α_{2}, α_{3}, λ_{1} is a chain and so l(α_{4}) = l(λ_{0}) ≤ l(λ_{1}). If l(λ_{2}) ≤ l(λ_{3}) we can apply Theorem 2.12 since κ_{2,3} is a shortest arc. Otherwise l(λ_{3}) ≤ l(λ_{2}) ≤ l(λ_{0}) and we can again apply Theorem 2.12 since κ_{3,0} = α_{1} is a shortest arc.

Now j_{3} = 4. We know that b_{1}, b_{2} ∈ 𝒪′_{c} since β_{1} does not cross Υ_{i_{3},4} . Consider i_{3} = 2. If a_{1} ∈ {b_{1}, b_{2}} then 𝒜′_{2} = α_{1}, β_{1} is a chain and α_{2} is a shortest arc. So a_{1} ∉ {b_{1}, b_{2}}. Suppose (a_{2}, a_{3}) = (b_{3}, b_{5}). If a_{1} = b_{4} then we can apply Theorem 2.6 and Lemma 2.3(ii) to contradict α_{1} = β_{3} being a shortest arc; see the first wire-frame, Figure 8. So a_{1} = c and we have that ℬ′_{3} = β_{1}, β_{2}, α_{1} , 𝒜′_{2} = α_{1}, β_{3} are both chains and so α_{2} is again a shortest arc. Next (a_{2}, a_{3}) = (b_{5}, b_{3}). We may suppose that a_{1} ≠ b_{4} . So a_{1} = c and 𝒜_{2}, ℬ is equivalent to a minimal chain pair on ℰ ; see the fifth wire-frame, Figure 12.

Next i_{3} = 3. First (a_{3}, a_{4}) = (b_{3}, b_{5}). If {a_{1}, a_{2}} ∌ b_{4} then, as α_{1} does not cross Υ_{3,4} it follows that a_{1}, a_{2} ∈ 𝒪′_{c}, i.e. {a_{1}, a_{2}} = {b_{1}, b_{2}} and 𝒜_{3}, ℬ is equivalent to a minimal chain pair on ℰ ; see the sixth wire-frame, Figure 12. So {a_{1}, a_{2}} ∋ b_{4} . Suppose a_{1} = b_{4} . If a_{2} = b_{1} then (a_{2}, a_{3}) = (b_{1}, b_{3}), a_{1} ≠ b_{2} and so α_{2}∪β_{1}∪β_{2} is a bracelet of shortest arcs. If a_{2} = b_{2} then we argue as follows; see the second wire-frame, Figure 8. Set α_{3} = κ_{3,4} , β_{3} = λ_{0} , β_{4} = κ_{3,5} , β_{2} = α_{2} = κ_{1,4} , α_{1} = κ_{1,5} which extends uniquely to K ∪ Λ. Each one of 𝒜′_{3} = α_{1}, α_{2}, κ_{0,4} , ℬ′_{3} = β_{1}, β_{2}, λ_{2} , ℬ′_{4} = β_{1}, β_{2}, β_{3}, κ_{0,5} is a chain so the hypotheses of Theorem 2.6 are satisfied: l(λ_{0}) = l(λ_{2}). Now 𝒜′_{2} = α_{1}, β_{1} is a chain, so α_{2} = κ_{1,4} is a shortest arc. Now α_{1} = κ_{1,5}, β_{1} = κ_{1,2} are also shortest arcs so, by Theorem 2.7, 𝒪 is the octahedral orbifold. So a_{2} ∉ {b_{1}, b_{2}} and each one of ℬ′_{4} = β_{1}, β_{2}, β_{3}, α_{1} , 𝒜′_{2} = α_{1}, β_{4} , ℬ′_{3} = β_{1}, β_{2}, α_{2} are chains: β_{3} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs. Next suppose a_{2} = b_{4} . If a_{1} = c and we can apply Theorem 2.7; see the third wire-frame. If a_{1} = b_{1} then ℬ′_{2} = β_{1}, α_{1} , 𝒜′_{3} = α_{1}, α_{2}, β_{2} are chains and so β_{2}, α_{3} are shortest arcs; we can apply Theorems 2.6 and 2.12 to show 𝒪 is 𝒪ct ;

Figure 8. Configurations of 𝒜_{i_{3}}, ℬ_{4}

see the fourth wire-frame. If a_{1} = b_{2} then (a_{1}, a_{3}) = (b_{2}, b_{3}), a_{2} ≠ b_{1} and so β_{2} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs.

Next (a_{3}, a_{4}) = (b_{5}, b_{3}). We may suppose that a_{2} ≠ b_{4} . Suppose a_{1} = b_{4} . For a_{2} ∉ {b_{1}, b_{2}} we have that both of ℬ′_{4} = β_{1}, β_{2}, β_{3}, α_{1} , 𝒜′_{2} = α_{1}, β_{4} are chains and so β_{4} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs. So a_{2} ∈ {b_{1}, b_{2}}. If a_{2} = b_{2} then α_{1}, β_{1}, α_{2}, β_{2} have exactly one cone point in common and we can apply Lemma 2.11: 𝒪 is 𝒪ct . If a_{2} = b_{1} we can apply Theorems 2.6 and 2.12— the fifth (respectively sixth) wire-frame illustrates the argument for l(α_{3}) ≤ l(β_{4}) (respectively l(β_{4}) ≤ l(α_{3})).

So a_{1} ≠ b_{4} and hence {a_{1}, a_{2}} = {b_{1}, b_{2}}; see the seventh and eighth wireframes. Set α_{3} = κ_{3,4} , β_{3} = κ_{3,5} , β_{4} = λ_{0} , β_{2} = κ_{2,3} which extends to K ∪ Λ. First 𝒜′_{3} = α_{1}, α_{2}, κ_{0,4} is a chain and so l(κ_{3,4}) ≤ l(κ_{0,4}). Likewise ℬ′_{3} = β_{1}, β_{2}, α_{3} , 𝒜′_{3} = α_{1}, α_{2}, β_{4} , ℬ′_{4} = β_{1}, β_{2}, β_{3}, κ_{0,5} are all chains so l(κ_{3,5}) = l(β_{3}) ≤ l(α_{3}) ≤ l(β_{4}) ≤ l(κ_{0,5}). Also ℬ′_{2} = β_{1}, κ_{2,4} is chain and so l(κ_{2,3}) ≤ l(κ_{2,4}). Again ℬ′_{4} = β_{1}, β_{2}, β_{3}, κ_{0,5} is a chain and so l(λ_{0}) ≤ l(κ_{0,5}). By Theorem 2.13 either l(κ_{0,4}) < l(κ_{3,4}) or l(κ_{0,5}) < l(κ_{3,5}) or l(κ_{2,4}) < l(κ_{2,3}) or l(κ_{0,5}) < l(λ_{0}).

Finally i_{3} = 4. First (a_{4}, a_{5}) = (b_{3}, b_{5}). Suppose a_{1} = b_{4} . Since α_{2} does not cross Υ_{4,4} it follows that a_{2}, a_{3} ∈ 𝒪′_{c} and hence {a_{2}, a_{3}} = {b_{1}, b_{2}}. Suppose (a_{2}, a_{3}) = (b_{1}, b_{2}); see the ninth wire-frame. We argue as follows. Set α_{4} = κ_{3,4} , β_{3} = λ_{0} , β_{4} = κ_{3,5} , α_{3} = β_{2} = κ_{1,4} which extends to K_{3} ∪ Λ_{3} ∪ κ_{1,4} ∪ κ_{2,5} . We have that 𝒜′_{4} = α_{1}, α_{2}, α_{3}, κ_{0,4} , ℬ′_{3} = β_{1}, β_{2}, λ_{2} , ℬ′_{4} = β_{1}, β_{2}, β_{3}, κ_{0,5} are all chains and so l(κ_{3,l}) = l(κ_{0,l}), l(λ_{0}) = l(λ_{2}) by Theorem 2.6. Now ℬ′_{2} = β_{1}, α_{1} is a chain and so β_{2} = κ_{1,4} is a shortest arc. Since α_{1} = κ_{2,5} is also a shortest arc we have l(κ_{1,4}) = l(κ_{2,5}). By Lemma 2.8, l(κ_{3,4}) = l(κ_{0,5}). Now ℬ′_{3} = β_{1}, β_{2}, κ_{0,4} is a chain and so l(λ_{0}) ≤ l(κ_{0,4}). Likewise β_{1} = κ_{1,2} is a shortest arc and so l(κ_{1,2}) ≤ l(κ_{3,0}). By Theorem 2.14, l(κ_{1,4}) = l(κ_{2,5}) > l(κ_{1,2}) which contradicts κ_{2,5} = α_{1} , κ_{1,4} = β_{2} being shortest arcs. If (a_{2}, a_{3}) = (b_{2}, b_{1}) we can apply Theorems 2.6 and 2.7; see the tenth wire-frame.

Next a_{2} = b_{4} . If {a_{1}, a_{3}} = {b_{1}, b_{2}} then β_{1} ∪α_{1} ∪α_{2} is a bracelet of shortest arcs. For either a_{1} ∉ {b_{1}, b_{2}} or a_{2} ∉ {b_{1}, b_{2}} and we can apply Theorems 2.6 and 2.12; see the 11–14th wire-frames. For a_{3} = b_{4} we have that {a_{1}, a_{2}} = {b_{1}, b_{2}} since α_{1} does not cross Υ_{4,4} . Here 𝒜 , ℬ is equivalent to a minimal chain pair on ℰ ; see the seventh and eighth wire-frames, Figure 12.

To finish (a_{4}, a_{5}) = (b_{5}, b_{3}). We may suppose that a_{3} ≠ b_{4} . For a_{1} = b_{4} we again have that {a_{2}, a_{3}} = {b_{1}, b_{2}}. Suppose (a_{2}, a_{3}) = (b_{2}, b_{1}). We can apply Theorems 2.6 and 2.12—the 15th (respectively 16th) wire-frame illustrates the argument for l(β_{3}) ≤ l(β_{4}) (respectively l(β_{4}) ≤ l(β_{3})). Similarly for (a_{2}, a_{3}) = (b_{1}, b_{2}). Suppose a_{2} = b_{4} . If {a_{1}, a_{3}} = {b_{1}, b_{2}} then β_{1} ∪ α_{1} ∪ α_{2} is a bracelet of shortest arcs. So a_{1} = c, a_{3} ∈ {b_{1}, b_{2}}—see the 17th and 18th wire-frames—or a_{1} ∈ {b_{1}, b_{2}}, a_{3} = c—see the 19th and 2Oth wire-frames—and we can apply Theorems 2.6 and 2.12.

Lemma 2.9. If l(κ_{3,0}) = l(κ_{1,4}), l(κ_{0,5}) = l(κ_{3,5}) then there exists an arc γ between either c_{2}, c_{1} or c_{2}, c_{4} such that l(γ) ≤ l(κ_{0,5}) = l(κ_{3,5}).

Proof. Let Γ = κ_{3,0} ∪ κ_{1,4} ∪ κ_{2,5} , a disjoint triple of arcs. As l(κ_{3,0}) = l(κ_{1,4}) the pair of pants 𝒪 \ Γ has rotational symmetry R exchanging κ_{3,0} , κ_{1,4} and fixing κ_{2,5} setwise.

The cone points c_{0} , c_{3} , c_{5} span an isosceles triangle ℐ bounded by κ_{3,0} ∪ κ_{3,5} ∪ κ_{0,5} . So R(c_{0}), R(c_{3}), R(c_{5}) spans an isometric isosceles triangle R(ℐ). As R(c_{0}), R(c_{3}) divides κ_{1,4} into equal length subarcs, either c_{1} or c_{4} lies on the κ_{1,4} edge of R(ℐ). If γ denotes the arc between this cone point and R(c_{5}) = c_{2} contained in R(ℐ) then γ has the required properties.

Suppose a minimal chain pair 𝒜 , ℬ such that (a_{2}, a_{3}) = (b_{1}, b_{3}), a_{1} ≠ b_{2} . Since a_{2} = b_{1} but a_{1} ≠ b_{2} : 𝒜′_{2} = α_{1}, β_{1} , ℬ′_{2} = β_{1}, α_{1} are both chains and so α_{2}, β_{2} are shortest arcs. Now (a_{2}, a_{3}) = (b_{1}, b_{3}) and so α_{2} ∪ β_{1} ∪ β_{2} is a bracelet of shortest arcs. By Proposition 2.10, 𝒜 , ℬ is equivalent to a minimal standard chain pair on 𝒪ct and no length 3 bracelet of links on 𝒪ct cuts off a cone point. Likewise if {a_{1}, a_{2}} = {b_{1}, b_{3}}. (For applications to the proof of Theorem 1.1 we note that no minimal chains on 𝒪 cross.)

Proposition 2.10. Suppose an orbifold 𝒪 has a length 3 bracelet of shortest arcs. Any minimal chain pair on 𝒪 is equivalent to a minimal chain pair on the 𝒪ct .

Proof. Let Υ be a length 3 bracelet of shortest arcs on 𝒪. Suppose Υ cuts off a cone point. By an application of Theorem 2.6 and then Lemma 2.3(ii), as in the proof of Proposition 2.4, we derive a contradiction. So Υ bounds a triangle. There exists Υ′ another length 3 bracelet of shortest arcs disjoint from Υ that bounds a triangle. Moreover, the conformal symmetry group of 𝒪 contains a subgroup isomorphic to S_{3} . This is a well-known result; see for example Schmutz [23, Lemma 5.1]. We will sketch the following proof of the existence of Υ′, along the lines of the proof of Lemma 2.9.

Relabel the arcs in Υ by κ_{3,0}, κ_{0,5}, κ_{3,5} ; label the shortest arc disjoint from Υ by κ_{1,4} ; and label the unique arc disjoint from κ_{3,0} ∪ κ_{1,4} by κ_{2,5} . Label the cone points so that κ_{k,l} is between c_{k}, c_{l}, et cetera. Again, let Γ = κ_{3,0}∪κ_{1,4}∪κ_{2,5} and so 𝒪 \Γ is a pair of pants. Denote by P_{3,0} (respectively P_{1,4} ) the common perpendicular between boundary components κ_{2,5} and κ_{3,0} (respectively κ_{2,5} and κ_{1,4} ); denote by D_{5} (respectively D_{2} ) a disc of radius l(κ_{3,0}) (respectively l(κ_{1,4})) about c_{5} (respectively c_{2} ); and denote by I_{3,0} (respectively I_{1,4} ) the interval of the boundary component κ_{3,0} (respectively κ_{1,4} ) inside D_{5} (respectively D_{2} ). So I_{3,0} , I_{1,4} are chords to D_{5} , D_{2} respectively. Let θ_{5} (respectively θ_{2} ) denote the angle that I_{3,0} (respectively I_{1,4} ) subtends at c_{5} (respectively c_{2} ). Finally, denote by θ_{3,0} (respectively θ_{1,4} ) the angle of an equilateral triangle of edge length l(κ_{3,0}) (respectively l(κ_{1,4})).

Since l(κ_{3,0}) ≤ l(κ_{1,4}) it follows that θ_{3,0} ≥ θ_{1,4} and that l(P_{3,0}) ≥ l(P_{1,4}). Moreover l(κ_{3,0}) ≤ l(κ_{1,4}), l(P_{3,0}) ≥ l(P_{1,4}) implies that θ_{5} ≤ θ_{2} . Since I_{1,4} subtends an angle θ_{2} ≥ θ_{5} = θ_{3,0} ≥ θ_{1,4} it follows that l(I_{1,4}) ≥ l(κ_{1,4}).

Unless l(I_{1,4}) = l(κ_{1,4}) and the cone points c_{1}, c_{4} lie at the ends of I_{1,4} , there exists an arc γ between c_{1}, c_{2} or c_{2}, c_{4} such that l(γ) < l(κ_{1,4}). This would contradict the shortness assumption on κ_{1,4} . So l(I_{1,4}) = l(κ_{1,4}) and c_{1}, c_{4} lie at the ends of I_{1,4} . So c_{2} , c_{1} , c_{4} span a bracelet of shortest arcs Υ′.

Let Ξ be the set of shortest arcs between some cone point on Υ and some cone point on Υ′. Using cut-and-paste arguments, arcs in Ξ do not cross either: each other or arcs in Υ ∪ Υ′. By the 𝒮_{3} action of the symmetry group: |Ξ| = 3 or 6. It is not hard to show that any minimal chain on 𝒪 must lie in Υ ∪ Υ′ ∪ Ξ. Moreover Υ ∪ Υ′ ∪ Ξ is graph isomorphic to a subgraph of the set of shortest arcs on 𝒪ct .

Suppose a minimal chain pair 𝒜 , ℬ is such that a_{2} = b_{2} and a_{1} , a_{3} , b_{1} and b_{3} are all distinct. We have that 𝒜′_{2} = α_{1}, β_{1} , ℬ′_{2} = β_{1}, α_{1} are both chains and so α_{2}, β_{2} are shortest arcs. It follows that α_{1}, β_{1}, α_{2}, β_{2} are four distinct shortest arcs with exactly one cone point in common. By Lemma 2.11, 𝒪 is the octahedral orbifold.

Lemma 2.11. Suppose, for some k, that κ_{k−1,k}, κ_{k,k+1} , κ_{k,l} are all shortest arcs for l = 4, 5. Then 𝒪 is the octahedral orbifold.

Proof. This result appeared as Lemma 5.2 in Schmutz [23]. However we offer the following proof. It illustrates the applications of Theorems 2.6 and 2.12 in the proof of Proposition 2.4. The arc set κ_{k−1,k}, κ_{k,k+1} , κ_{k,l} for l = 4, 5 extends uniquely to the arc set K ∪ Λ.

Suppose l(λ_{k+1}) ≤ l(λ_{k−2}). We know that l(κ_{k,l}) ≤ l(κ_{k+1,l}) for l = 4, 5 and so l(λ_{k−1}) ≤ l(λ_{k+1}) by Theorem 2.6. So l(λ_{k−1}) ≤ l(λ_{k+1}) ≤ l(λ_{k−2}). If l(λ_{k−1}) ≤ l(λ_{k}) then we can apply Theorem 2.12. If l(λ_{k}) ≤ l(λ_{k−1}) then l(λ_{k}) ≤ l(λ_{k−1}) ≤ l(λ_{k+1}) and we can again apply Theorem 2.12.

Suppose l(λ_{k−2}) ≤ l(λ_{k+1}). We know that l(κ_{k,l}) ≤ l(κ_{k−1,l}) for l = 4, 5 and so l(λ_{k}) ≤ l(λ_{k−2}) by Theorem 2.6. So l(λ_{k}) ≤ l(λ_{k−2}) ≤ l(λ_{k+1}). If l(λ_{k}) ≤ l(λ_{k−1}) then we can apply Theorem 2.12. If l(λ_{k−1}) ≤ l(λ_{k}) then l(λ_{k−1}) ≤ l(λ_{k}) ≤ l(λ_{k−2}) and we can again apply Theorem 2.12.

Theorem 2.12. Suppose, for some k, that κ_{k,k+1} is a shortest arc and l(λ_{k}) ≤ min{l(λ_{k−1}), l(λ_{k+1})} then 𝒪 is the octahedral orbifold.

Proof. Suppose k = 3. Cut 𝒪_{3,0} open along κ_{0,5} , κ_{3,4} so as to obtain a simply connected domain Ω. Take a lift of Ω to the universal cover 𝒪̃ and, without confusion, give the geodesics having non-trivial intersection with Ω the same labels as on 𝒪. So Ω is bounded by κ_{0,5} , λ_{0} , κ_{3,4} , λ_{2} . In the same cyclic order label the orbits of cone points that lie on the boundary of Ω by c_{5} , c_{0} , c′_{5} , c′_{4} , c_{3} and c_{4} . The point c_{0} (respectively c_{3} ) lies at the midpoint of the κ_{0,5} edge (respectively κ_{3,4} edge). By inspection λ_{3} is the diagonal of Ω between vertices c_{5} and c′_{4} . As κ_{3,0} is a shortest arc c_{5} , c′_{4} (and hence c′_{5} , c_{4} ) cannot lie inside D_{0} ∪ D_{3} , where D_{k} denotes the closed disc of radius κ_{3,0} about c_{k}.

To prove the result we will show that either: l(λ_{3}) > l(λ_{2}) or l(λ_{3}) > l(λ_{0}) or l(λ_{3}) = l(λ_{2}) = l(λ_{0}) and 𝒪 is the octahedral orbifold. Let θ denote the angle in an equilateral triangle of edge length l(κ_{3,0}). The maximal length of a shortest arc on an orbifold 𝒪 occurs exactly for 𝒪 the octahedral orbifold; see Näätänen [22] or Schmutz [23]. It is a simple consequence that θ ≥ π/4 with equality if and only if 𝒪 is the octahedral orbifold.

Claim. Either ^{}c̸_{5}c_{0}c′_{4} ≥ 2θ or ^{}c̸_{5}c_{3}c′_{4} ≥ 2θ.

Now ^{}c̸_{5}c_{0}c′_{4} > π/2 implies ^{}c̸′_{5}c_{0}c′_{4} < π/2 and hence that l(λ_{0}) < l(λ_{3}). Likewise ^{}c̸_{5}c_{3}c′_{4} > π/2 implies ^{}c̸_{5}c_{3}c_{4} < π/2 and hence that l(λ_{2}) < l(λ_{3}). So we must have ^{}c̸_{5}c_{0}c′_{4} = ^{}c̸_{5}c_{3}c′_{4} = 2θ = π/2. That is 𝒪 is the octahedral orbifold.

Proof of the claim. Since c_{5} lies outside D_{0} ∪ D_{3} either ^{}c̸_{5}c_{0}c_{3} ≥ θ or c̸_{5}c_{3}c_{0} ≥ θ. Likewise since c′_{4} lies outside D_{0} ∪ D_{3} either ^{}c̸′_{4}c_{0}c_{3} ≥ θ or c̸′_{4}c_{3}c_{0} ≥ θ. If ^{}c̸_{5}c_{0}c_{3} ≥ θ and ^{}c̸′_{4}c_{0}c_{3} ≥ θ then ^{}c̸_{5}c_{0}c′_{4} ≥ 2θ. Likewise, if c̸_{5}c_{3}c_{0} ≥ θ and ^{}c̸′_{4}c_{3}c_{0} ≥ θ then ^{}c̸_{5}c_{3}c′_{4} ≥ 2θ. So, up to relabelling, we may suppose that ^{}c̸_{5}c_{3}c_{0} ≤ θ and ^{}c̸′_{4}c_{0}c_{3} ≤ θ as in Figure 9. Let θ_{5} = ^{}c̸_{5}c_{0}c_{3} ≥ θ and θ_{4} = ^{}c̸′_{4}c_{3}c_{0} ≥ θ. If θ_{5} ≥ 2θ or θ_{4} ≥ 2θ we are done. So suppose not.

Consider points x_{5} , x′_{4} , to the c_{5}, c′_{4} side of κ_{3,0} respectively, such that x̸_{5}c_{0}c_{3} = θ_{5} and ^{}x̸_{5}c_{3}c_{0} = 2θ − θ_{5} ; and ^{}x̸′_{4}c_{3}c_{0} = θ_{4} and ^{}x̸′_{4}c_{0}c_{3} = 2θ − θ_{4} . Next we show that x_{5} ∈ D_{0} . (Similarly x′_{4} ∈ D_{3} ).

We have constructed the points A, E, C and C′ such that ^{}A̸BC = B̸AC = ^{}A̸CB = θ and ^{}B̸AC′ = θ_{5} , ^{}A̸BC′ = 2θ − θ_{5} . We let O denote the intersection of AC, BC′. Now ^{}O̸AC′ = ^{}O̸BC = θ_{5} − θ and ^{}A̸OC′ = ^{}B̸OC and OA is shorter than OB. So comparing the triangles OAC′ and OBC we see that AC′ is shorter than BC and we are done.

Now ^{}c̸_{5}c_{0}c′_{4} ≥ ^{}c̸_{5}c_{0}x′_{4} = 2θ+(θ_{5}−θ_{4}), ^{}c̸_{5}c_{3}c′_{4} ≥ ^{}x̸_{5}c_{3}c′_{4} = 2θ+(θ_{4}−θ_{5}).

Figure 9. Arcs and lift of Ω in Theorem 2.12

Theorem 2.13. Either l(κ_{0,4}) < l(κ_{3,4}) or l(κ_{0,5}) < l(κ_{3,5}) or l(κ_{2,4}) < l(κ_{2,3}) or l(κ_{0,5}) < l(λ_{0}).

Proof. We suppose that l(κ_{3,4}) ≤ l(κ_{0,4}), l(κ_{3,5}) ≤ l(κ_{0,5}), l(κ_{2,3}) ≤ l(κ_{2,4}) and show that l(κ_{0,5}) < l(λ_{0}). First, by Theorem 2.6, l(λ_{2}) ≤ l(λ_{0}). Also ⋃_{l=4,5} κ_{2,l} ∪ κ_{3,l} bound the quadrilateral spanned by λ_{2} ∪ κ_{2,3} . By Lemma 3.3, l(κ_{2,4}) < l(κ_{2,3}) or l(κ_{3,5}) < l(λ_{2}). So l(κ_{3,5}) < l(λ_{2}).

Cut A_{3,0} open along κ_{3,5} to obtain a simply connected domain Ω. Choose a lift of Ω in the universal cover of the annulus 𝒜_{3,0} . Label the geodesics around the boundary of Ω by κ_{3,0} , κ_{3,5} , λ_{2} , λ_{0} and κ′_{3,5} , in cyclic order. Without confusion, give the lifts of κ_{0,4} , κ_{0,5} and κ_{3,4} having non-trivial intersection with Ω the same labels. In the same cyclic order, label orbits of cone points around the boundary of Ω: c_{0} , c_{3} , c_{5} , c_{4} , c′_{5} and c′_{3} . Let P_{5,3} , P_{4,3} and P′_{5,3} denote the perpendiculars to κ_{3,0} from c_{5} , c_{4} , c′_{5} and let f_{5,3} = P_{5,3}∩κ_{3,0} , f_{4,3} = P_{4,3}∩κ_{3,0} , f′_{5,3} = P′_{5,3} ∩ κ_{3,0} . By choosing orientation we may suppose that P_{5,3} is to the left and P′_{5,3} to the right of P_{4,3} .

We now observe that c_{0} must lie strictly between P_{4,3} and P′_{5,3} . If c_{0} is left of P_{4,3} then l(κ_{0,4}) < l(κ_{3,4}) and if c_{0} is to the right of P′_{5,3} then l(κ_{0,5}) < l(κ_{3,5}).

If c_{3} = f_{5,3} set η = π/2, if c_{3} = f_{4,3} set ν = π/2, ϕ = 0. Otherwise we label angles as follows. Set θ = ^{}c̸_{5}c_{3}c_{4} , θ′ = ^{}c̸′_{5}c_{0}c_{4} and ψ = ^{}c̸_{5}c_{4}c_{3} , ψ′ = ^{}c̸′_{5}c_{4}c_{0} and ϕ = ^{}c̸_{3}c_{4}f_{4,3} , ϕ′ = ^{}c̸_{0}c_{4}f_{4,3} and ν = ^{}c̸_{4}c_{3}f_{4,3} , ν′ = ^{}c̸_{4}c_{0}f_{4,3} and η = ^{}c̸_{5}c_{3}f_{5,3} , η′ = ^{}c̸′_{5}c_{0}f′_{5,3} ; see Figure 10.

Since l(κ_{3}, l) ≤ l(κ_{0,l}) for l = 4, 5 we have that ν ≥ ν′, η ≥ η′. If c_{3} is between P_{5,3} and P_{4,3} then θ = π − (ν + η). If c_{3} lies strictly to the left of P_{5,3} then θ = η − ν < (π − η) − ν since η < π/2. If c_{3} lies strictly to the right of P_{4,3} then θ = ν − η < (π − ν) − η since ν < π/2. In each case θ ≤ π − (ν + η) ≤ π − (ν′ + η′) = θ′.

Also l(κ_{3,4}) ≤ l(κ_{0,4}) implies that ϕ ≤ ϕ′. Since l(λ_{2}) ≤ l(λ_{0}), by comparing the birectangles 𝒬_{2,3}, 𝒬_{0,3} , we have that ϕ + ψ ≥ ϕ′ + ψ′. It follows that ψ ≥ ψ′. Now l(κ_{3,5}) < l(λ_{2}) and so θ > ψ. Therefore θ′ ≥ θ > ψ ≥ ψ′: l(κ_{0,5}) < l(λ_{0}).

Figure 10. c_{3} between P_{5,3} , P_{4,3} and c_{3} to the left of P_{5,3}

Theorem 2.14. If l(κ_{3,4}) = l(κ_{3,5}) = l(κ_{0,4}) = l(κ_{0,5}) and l(κ_{1,4}) = l(κ_{2,5}), l(λ_{0}) = l(λ_{2}) ≤ l(κ_{0,4}) and l(κ_{1,2}) ≤ l(κ_{3,0}) then l(κ_{1,4}) = l(κ_{2,5}) > l(κ_{1,2}).

Proof. By Lemma 2.8 both annuli A_{1,2}, A_{3,0} have rotational symmetry exchanging λ_{0} ↔ λ_{2} . It follows that 𝒪 \ κ_{1,2} ∪ κ_{3,0} has rotational symmetry exchanging λ_{0} ↔ λ_{2} . So κ_{1,2} ∪ P_{4,1} ∪ P_{4,3} ∪ κ_{3,0} ∪ P_{5,3} ∪ P_{5,1} divides 𝒪 into a pair of isometric right hexagons, ℋ_{0} ⊃ λ_{0} , ℋ_{2} ⊃ λ_{2} .

Consider ℋ_{0} . First we note that c_{0} lies at the midpoint of the κ_{3,0} edge of ℋ_{0} . The common perpendicular between the κ_{1,2} , κ_{3,0} edges of ℋ_{0} divides ℋ_{0} into a mirror pair of right pentagons 𝒫_{l} where 𝒫_{l} has c_{l} as a vertex. Relabel the edges of 𝒫_{4} : A, …, E in cyclic order such that A = l(P_{4,1}), …, E = l(P_{4,3}). Label by F the diagonal from A∩E to D∩C; by G the perpendicular from A∩E to C; and by H the diagonal from A∩E to C ∩B. Label by θ the angle between A, H; ψ the angle between G, H; and ϕ the angle between C, H. We observe that l(κ_{1,4}) ≥ l(P_{4,1}) = A. To complete the proof we show that A > 2B = l(κ_{1,2}).

Consider the triangle ABH:

sinh H = \frac{sinh A}{sin(π/2 − ϕ)}  or  sinh H = \frac{sinh A}{cos ϕ}.

Now consider the double of ABH along A:

\frac{sinh 2B}{sin 2θ} = \frac{sinh H}{sin(π/2 − ϕ)}  or  \frac{sinh 2B}{sinh H} = \frac{sin 2θ}{cos ϕ}

and so we have

\frac{sinh 2B}{sinh A} = \frac{sin 2θ}{cos^{2} ϕ}.

Now B = l(κ_{1,2})/2 ≤ l(κ_{3,0})/2 = D implies that 0 < θ + ψ ≤ π/4. So 0 < 2θ ≤ π/2 − 2ψ < π/2. Therefore sin 2θ ≤ sin(π/2 − 2ψ).

Likewise B ≤ D implies that H ≥ F. Moreover F = l(κ_{0,4}) ≥ l(λ_{0}) = 2G. Consider CGH doubled along C. As H ≥ 2G we have that 2ϕ ≤ ψ. Since ψ < π/4 it follows that 0 < π/4 − ψ ≤ π/4 − 2ϕ < π/4. So 0 < π/2 − 2ψ ≤ π/2 − 4ϕ < π/2, and hence sin(π/2 − 2ψ) ≤ sin(π/2 − 4ϕ). Therefore sin 2θ ≤ sin(π/2 − 2ψ) ≤ sin(π/2 − 4ϕ) = cos 4ϕ and so

\frac{sinh 2B}{sinh A} ≤ \frac{cos 4ϕ}{cos^{2} ϕ}.

Now cos^{2} ϕ−cos 4ϕ = cos^{2} ϕ−2(2 cos^{2} ϕ−1)^{2}−1 = (8 cos^{2} ϕ−1)(1−cos^{2} ϕ). Since 0 < ϕ < π/8 it follows that cos^{2} ϕ > cos 4ϕ. So sinh A > sinh 2B, A > 2B.

Figure 11. The pentagon 𝒫_{4}

Proof of Proposition 2.5. Suppose that a minimal chain pair 𝒜_{i_{3}}, ℬ_{j_{3}} satisfies the hypotheses of Proposition 2.4. We need to show that any minimal standard chain pair 𝒜 , ℬ containing 𝒜_{i_{3}}, ℬ_{j_{3}} is equivalent to a minimal standard chain pair on ℰ . Throughout, Proposition 2.1 will be used without mention.

Consider (i_{3}, j_{3}) = (2, 3). By Proposition 2.4, (a_{2}, a_{3}) = (b_{2}, b_{4}) and a_{1} = b_{1} = c. Set α_{2} = κ_{3,4} , β_{2} = κ_{3,5} , β_{3} = λ_{0} which extends to K_{3} ∪ Λ_{3} . Each one of 𝒜′_{2} = α_{1}, κ_{0,4} , ℬ′_{2} = β_{1}, κ_{0,5} , ℬ′_{3} = β_{1}, β_{2}, λ_{2} is a chain and so l(λ_{0}) = l(λ_{2}) by Theorem 2.6. Consider the possibilities for β_{4}, α_{3}, α_{4} . Unless Υ_{3,4} = β_{3} ∪ α_{3} ∪ α_{4} is a bracelet, 𝒜 , ℬ is equivalent to a minimal standard chain pair on ℰ .

Suppose Υ_{3,4} is a bracelet. Exchange labels α_{∗} ↔ β_{∗} and set α′_{3} = λ_{2} . Suppose Υ_{3,4} = α_{3} ∪ β_{3} ∪ β_{4} cuts off a cone point. Then (a_{3}, a_{4}) = (b_{5}, b_{3}) with {a_{1}, a_{2}} ≠ b_{4} and by applying Lemma 3.3 and Theorem 2.13 we can derive a contradiction. If Υ_{3,4} = α_{3} ∪ β_{3} ∪ β_{4} bounds a triangle, then Υ′_{3,4} = α′_{3} ∪ β_{3} ∪ β_{4} cuts off a cone point. The same argument can now be applied, where we use the fact that l(α′_{3}) = l(α_{3}).

Next, consider (i_{3}, j_{3}) = (2, 4). Again, by Proposition 2.4, (a_{2}, a_{3}) = (b_{5}, b_{3}) and c = α_{1} ∉ {b_{1}, b_{2}}. Set α_{2} = κ_{3,4} , β_{4} = κ_{3,5} , β_{3} = λ_{0} which again extends to K_{3} ∪ Λ_{3} . Again by Theorem 2.6, l(λ_{0}) = l(λ_{2}). Now we consider α_{3}, α_{4} . Unless Υ_{3,4} = β_{3} ∪ α_{3} ∪ α_{4} is a bracelet, 𝒜 , ℬ is equivalent to a standard minimal chain on ℰ . Suppose Υ_{3,4} is a bracelet. Now both of 𝒜′_{4} = α_{1}, α_{2}, α_{3}, β_{1} , ℬ′_{2} = β_{1}, α_{4} are chains and so α_{4}, β_{2} are shortest arcs. If α_{3} ≠ β_{2} then 𝒜′_{3} = α_{1}, α_{2}, β_{2} is a chain and so α_{3} is a shortest arc. In either case β_{1}, α_{3}, α_{4} are all shortest arcs and l(λ_{0}) = l(λ_{2}). By Theorem 2.7, 𝒪 is the octahedral orbifold. As in the proof of Proposition 2.4 this gives a contradiction.

For (i_{3}, j_{3}) = (3, 3), (4, 3), (4, 4) there is nothing to prove. It remains to consider (i_{3}, j_{3}) = (3, 4). By Proposition 2.4 we have that (a_{3}, a_{4}) = (b_{3}, b_{5}) and {a_{1}, a_{2}} = {b_{1}, b_{2}}. Set α_{3} = κ_{3,4} , β_{4} = κ_{3,5} , β_{3} = λ_{0} which extends to K_{3} ∪ Λ_{3} . We can apply Theorem 2.6: l(λ_{0}) = l(λ_{2}). Unless α_{4} = β_{4} , 𝒜 , ℬ is equivalent to a minimal standard chain pair on ℰ . Suppose α_{4} = β_{4} . Set β_{3} = κ̂_{3,4} , α_{4} = κ̂_{3,5} , α_{3} = λ̂_{0} which again extends to K̂_{3} ∪ Λ̂_{3} . Again we can apply Theorem 2.6: l(λ̂_{0}) = l(λ̂_{2}). The arcs λ_{2}, λ̂_{2} between b_{3}, b_{4} and a_{3}, a_{4} , respectively, cross in a single point and have an endpoint b_{3} = a_{3} in common. By one of the cut-and-paste arguments we use in the proof of Proposition 3.1 we have that l(α_{3}) < l(λ̂_{2}) or l(β_{3}) < l(λ_{2}) which gives a contradiction.

3. Proof of Theorem 1.1

As in the previous section we shall work on the quotient orbifold 𝒪 = 𝒮 /𝒥 . We consider pairs of crossing standard minimal chains 𝒜 , ℬ and derive contradictions. We say that α_{i_{1}}, β_{j_{1}} are the first crossing links if i_{1} = min i ∈ {1, …, 4} such that α_{i}, β_{j} cross, j_{1} = min j ∈ {1, …, 4} such that α_{i_{1}}, β_{j} cross. The crossing minimal chain pair 𝒜_{i_{1}}, ℬ_{j_{1}} has exactly one pair of crossing arcs: α_{i_{1}}, β_{j_{1}} . In Proposition 3.1 we show that |α_{i_{1}} ∩β_{j_{1}}| < 2. It remains to consider |α_{i_{1}} ∩β_{j_{1}}| = 1

Figure 12. Minimal chain pairs with j_{3} = 3, 4

i.e., α_{i_{1}}, β_{j_{1}} have distinct endpoints and a single crossing point −α_{i_{1}} ∪β_{j_{1}} has the form an ‘X ’ on 𝒪.

Proposition 3.1. We have |α_{i_{1}} ∩ β_{j_{1}}| < 2.

Proof. First we consider α_{i_{1}}, β_{j_{1}} having more than one crossing point. We then consider α_{i_{1}}, β_{j_{1}} having one crossing point and one or two endpoints in common.

On α_{i_{1}} label the crossing point nearest to its tail by a_{t} and the crossing point nearest to its head by a_{h}. Likewise for the crossing points on β_{j_{1}} . The points a_{t}, a_{h}, b_{t}, b_{h} divide α_{i_{1}} into either three, four or five subarcs. Label the subarc of α_{i_{1}} between a_{i_{1}} , a_{t} by α_{t}; between b_{t}, b_{h} by α_{m}; and between a_{h}, a_{i_{1}+1} by α_{h}. Likewise for the subarcs of β_{j_{1}} .

Let α′_{i_{1}} denote the geodesic in the endpoint fixed homotopy class of α_{t} ∪ β_{m} ∪ α_{h}. By its definition α′_{i_{1}} is simple. Let 𝒜_{i_{1}} = α_{1}, …, α′_{i_{1}} . Suppose α′_{i_{1}} crosses α_{i} for some 1 ≤ i < i_{1} . Choose lifts of α_{t}, β_{m}, α_{h}, α′_{i_{1}} in H^{2} so that α_{t} ∪β_{m} ∪α_{h} ∪α′_{i_{1}} bounds a rectangle (crossed or uncrossed). A lift of α_{i} intersects α′_{i_{1}} between a_{i_{1}}, a_{i_{1}+1} . This lift also intersects eitherα_{t} or β_{m} or α_{h}. This intersection projects to a crossing point of α_{i} with α_{i_{1}} or β_{j_{1}} on 𝒪 which gives a contradiction. So 𝒜′_{i_{1}} is a chain.

Let β′_{j_{1}} denote the geodesic in the endpoint fixed homotopy class of β_{t} ∪ α_{m} ∪ β_{h}. Again ℬ′_{j_{1}} = β_{1}, …, β′_{j_{1}} is a chain. We have that l(α′_{i_{1}}) < l(α_{i_{1}}) or l(β′_{j_{1}}) < l(β_{j_{1}}). If l(α_{m}) ≤ l(β_{m}) then l(β′_{j_{1}}) < l(β_{t}) + l(α_{m}) + l(β_{h}) ≤ l(β_{t}) + l(β_{m}) + l(β_{h}) ≤ l(β_{j_{1}}). If l(β_{m}) ≤ l(α_{m}) then l(α′_{i_{1}}) < l(α_{t}) + l(β_{m}) + l(α_{h}) ≤ l(α_{t}) + l(α_{m}) + l(α_{h}) ≤ l(α_{i_{1}}).

We now consider α_{i_{1}}, β_{j_{1}} having one crossing point and one or two endpoints in common. We will suppose that a_{i_{1}} = b_{j_{1}} (the other possibilities follow similarly). The crossing point divides α_{i_{1}} into exactly two subarcs which we label α_{t}, α_{h}, so that α_{t} ∋ α_{i_{1}} . Likewise for β_{j_{1}} . Let α′_{i_{1}}, β′_{j_{1}} denote the geodesic in the endpoint fixed homotopy class of β_{t} ∪α_{h}, α_{t} ∪β_{h} respectively. Again we have that 𝒜′_{i_{1}} = α_{1}, …, α_{i_{1}−1}, α′_{i_{1}} and ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, β′_{j_{1}} are both chains and that l(α′_{i_{1}}) < l(α_{i_{1}}) or l(β′_{j_{1}}) < l(β_{j_{1}}).

We say that 𝒜_{i_{1}}, ℬ_{j_{1}} is of type (I) if {a_{i_{1}}, a_{i_{1}+1}} = {b_{j}, b_{j+1}}, for some j < j_{1} − 1, up to relabelling α_{∗} ↔ β_{∗}. We say that 𝒜_{i_{1}}, ℬ_{j_{1}} is of type (II) if {a_{i_{1}}, a_{i_{1}+1}} = {b_{j−1}, b_{j+1}} or {a_{i_{1}−1}, a_{i_{1}+1}} = {b_{j}, b_{j+1}}, for some j < j_{1} − 1, up to relabelling α_{∗} ↔ β_{∗}.

Proposition 3.2. Type (I), (II) minimal chain pairs give a contradiction.

Proof. Suppose 𝒜_{i_{1}}, ℬ_{j_{1}} is such that {a_{i_{1}}, a_{i_{1}+1}} = {b_{j}, b_{j+1}} for some j < j_{1} − 1. So arcs α_{i_{1}}, β_{j} share endpoints. We know that these arcs do not cross and that β_{j_{1}} crosses α_{i_{1}} but does not cross β_{j}. So Γ_{i_{1},j} = α_{i_{1}} ∪ β_{j} must be a bracelet. As j < j_{1}−1 ≤ 3 it follows that j = 1 or j = 2 which by Proposition 2.1 gives a contradiction.

Suppose now that 𝒜_{i_{1}}, ℬ_{j_{1}} is such that {a_{i_{1}−1}, a_{i_{1}+1}} = {b_{j}, b_{j+1}} for some j < j_{1} − 1. The other possibility follows similarly. So Υ_{j,i_{1}} = β_{j} ∪ α_{i_{1}−1} ∪ α_{i_{1}} is a bracelet. The arc β_{j_{1}} crosses only α_{i_{1}} ⊂ Υ_{j,i_{1}} . The endpoints b_{j_{1}}, b_{j_{1}+1} of β_{j_{1}} lie off Υ_{j,i_{1}} ; b_{j_{1}}, b_{j_{1}+1} are distinct from b_{j}, b_{j+1} ∈ β_{j} since j < j_{1} − 1 and we know that b_{j_{1}}, b_{j_{1}+1} are both distinct from a_{i_{1}} , the other cone point on Υ_{j,i_{1}} . So b_{j_{1}}, b_{j_{1}+1} must lie in different components of 𝒪 \ Υ_{j,i_{1}} . As there is only one other cone point lying off Υ_{j,i_{1}} it follows that Υ_{j,i_{1}} cuts off b_{j_{1}} or b_{j_{1}+1} . By Proposition 2.4, 1 < j, 2 < i_{1} . Since j < j_{1} − 1 ≤ 3, i_{1} ≤ 4 it follows that (j, i_{1}) = (2, 3) or (j, i_{1}) = (2, 4). As 2 = j < j_{1} − 1 ≤ 3 we have that j_{1} = 4. So Υ_{2,3} or Υ_{2,4} cuts off b_{4} or b_{5} . By Proposition 2.4 this gives a contradiction.

Figure 13. Examples of (a)–(e) of Proposition 3.4

Let ε_{t,t} ∪ ε_{t,h} ∪ ε_{h,h} ∪ ε_{h,t} denote the bracelet that bounds the quadrilateral spanned by α_{i_{1}} ∪ β_{j_{1}} ; we label so that ε_{t,h} is between a_{i_{1}}, b_{j_{1}+1} and ε_{h,t} is between a_{i_{1}+1}, b_{j_{1}} , et cetera. As in the proof of Proposition 3.1 none of these arcs cross 𝒜_{i_{1}}, ℬ_{j_{1}} . Similarly l(ε_{t,t}) < l(α_{i_{1}}) or l(ε_{h,h}) < l(β_{j_{1}}) and all other such combinations. We have proved:

Lemma 3.3. Either l(ε_{t,t}) < l(α_{i_{1}}) or l(ε_{h,h}) < l(β_{j_{1}}), et cetera.

Proposition 3.4. Up to relabelling α_{∗} ↔ β_{∗}, we have a contradiction:

(a) If 𝒜′_{i_{1}} = α_{1}, …, α_{i_{1}−1}, ε_{t,h}, ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{h,t} are both chains.

Side-pairing elements of Maskit’s fundamental domain 33 (b) If either i_{1} = 1 or 𝒜′_{i_{1}} = a_{1}, …, α_{i_{1}−1}, β_{1} is a chain and either ℬ′_{j_{1}} =

β_{1}, …, β_{j_{1}−1}, ε_{t,t} or ℬ′′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{h,t} is a chain.

(c) If i_{1} = 2 and 𝒜′_{2} = α_{1}, ε_{h,h}, ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{t,t} are both chains.

(d) If i_{1} = 2, j_{1} = 4 and 𝒜′_{2} = α_{1}, β_{2} , ℬ′_{2} = β_{1}, α_{2} , ℬ′′_{2} = β_{1}, ε_{h,t} and ℬ′_{4} = β_{1}, β_{2}, β_{3}, ε_{t,h} are all chains.

(e) If for some j < j_{1} : 𝒜′_{i_{1}} = α_{1}, …, α_{i_{1}−1}, β_{j}, ℬ′_{j} = β_{1}, …, β_{j−1}, α_{i_{1}} , ℬ′′_{j} = β_{1}, …, β_{j−1}, ε_{t,h} and ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{h,t} are all chains.

Proof. For each part we have a contradiction by Lemma 3.3. For (a) we have l(α_{i_{1}}) ≤ l(ε_{t,h}), l(β_{j_{1}}) ≤ l(ε_{h,t}). For (b) suppose ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{t,t} is a chain. We then have l(α_{i_{1}}) = l(β_{1}) ≤ l(ε_{h,h}), l(β_{j_{1}}) ≤ l(ε_{t,t}). For (c) we have l(α_{2}) ≤ l(ε_{h,h}), l(β_{j_{1}}) ≤ l(ε_{t,t}). For (d) we have l(α_{2}) = l(β_{2}) ≤ l(ε_{h,t}), l(β_{4}) ≤ l(ε_{t,h}). For (e) we have l(α_{i_{1}}) = l(β_{j}) ≤ l(ε_{t,h}), l(β_{j_{1}}) ≤ l(ε_{h,t}).

Up to relabelling we may suppose that i_{1} ≤ j_{1} . Since α_{i_{1}}, β_{j_{1}} have distinct endpoints a_{i_{1}}, a_{i_{1}+1}, b_{j_{1}}, b_{j_{1}+1} are all distinct.

First i_{1} = 1. If {a_{1}, a_{2}} ⊄ {b_{1}, …, b_{j_{1}−1}} then either ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{t,t} or ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{h,t} is a chain and we can apply Proposition 3.4(b). So {a_{1}, a_{2}} ⊂ {b_{1}, …, b_{j_{1}−1}} and we have type (I) or (II). So i_{1} > 1. Suppose that b_{j_{1}+1} ∉ {a_{1}, …, a_{i_{1}−1}}, a_{i_{1}+1} ∉ {b_{1}, …, b_{j_{1}−1}}. We then have that 𝒜′_{i_{1}} = α_{1}, …, α_{i_{1}−1}, ε_{t,h}, ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{h,t} are both chains and we can apply Proposition 3.4(a). So we may suppose that either b_{j_{1}+1} ∈ {a_{1}, …, a_{i_{1}−1}} or a_{i_{1}+1} ∈ {b_{1}, …, b_{j_{1}−1}}.

Figure 14. Applications of Proposition 3.4 for i_{1} = 2

Next i_{1} = 2. First we suppose b_{j_{1}+1} = a_{1} . If a_{2} ∉ {b_{1}, …, b_{j_{1}−1}} then ℬ′_{j_{1}} = β_{1}, …, β_{j_{1}−1}, ε_{t,t} is a chain. We note that ε_{h,h} is between b_{j_{1}+1} = a_{1}, a_{3} and so 𝒜′_{2} = α_{1}, ε_{h,h} is also a chain and we can apply Proposition 3.4(c). So a_{2} ∈ {b_{1}, …, b_{j_{1}−1}}.

For j_{1} = 2: Proposition 3.4(b), see the first wire-frame, Figure 14. Suppose j_{1} = 3. If a_{3} ∈ {b_{1}, b_{2}} type (I). If a_{3} ∉ {b_{1}, b_{2}}: Proposition 3.4(b), see the second and third wire-frames. For j_{1} = 4 if a_{3} ∈ {b_{1}, b_{2}, b_{3}}: type (I) or type (II). So a_{3} = b_{6} . If a_{2} ∈ {b_{1}, b_{2}}: Proposition 3.4(b), see the fourth and fifth wireframes. If a_{2} = b_{3} then 𝒜′_{2} = α_{1}, β_{3} , ℬ′_{3} = β_{1}, β_{2}, α_{2} , ℬ′′_{3} = β_{1}, β_{2}, ε_{t,h} and ℬ′_{4} = β_{1}, β_{2}, β_{3}, ε_{h,t} are all chains, Proposition 3.4(e), see the sixth wire-frame.

Therefore a_{1} ≠ b_{j_{1}+1}, a_{3} ∈ {b_{1}, …, b_{j_{1}−1}}.

For j_{1} = 2, a_{1} ≠ b_{3} , a_{3} = b_{1} . If a_{1} = b_{2} , Proposition 3.4(b), see the seventh wire-frame. So a_{1} ∈ {b_{4}, b_{5}, b_{6}}, Proposition 3.4(c); see the eighth wire-frame.

For j_{1} = 3, a_{1} ≠ b_{4} , a_{3} ∈ {b_{1}, b_{2}}. If a_{2} ∈ {b_{1}, b_{2}}, type (I). So a_{2} ∈ {b_{5}, b_{6}}. If a_{1} ∈ {b_{1}, b_{2}}, type (II). If a_{1} = b_{3} , Proposition 3.4(b); see the ninth and tenth wire-frames. So a_{3} ∈ {b_{1}, b_{2}}, {a_{2}, a_{1}} = {b_{5}, b_{6}} and we can apply Theorem 3.7, see Figure 21.

For j_{1} = 4, a_{1} ≠ b_{5} , a_{3} ∈ {b_{1}, b_{2}, b_{3}}. If a_{2} ∈ {b_{1}, b_{2}, b_{3}}, type (I) or (II). So a_{2} = b_{6} . If a_{1} ∈ {b_{1}, b_{2}, b_{3}}, type (I) or (II) unless {a_{1}, a_{3}} = {b_{1}, b_{3}}. If (a_{1}, a_{3}) = (b_{1}, b_{3}), Proposition 3.4(b); see the eleventh wire-frame. If (a_{1}, a_{3}) = (b_{3}, b_{1}), Proposition 3.4(d); see the 12th wire-frame. So a_{1} = b_{4} , Proposition 3.4(b); see the 13th, 14th and 15th wire-frames.

Now i_{1} = 3. First j_{1} = 3. Up to relabelling we may suppose b_{4} ∈ {a_{1}, a_{2}}. If b_{3} ∈ {a_{1}, a_{2}}, type (I). So b_{3} ∈ {a_{5}, a_{6}}. Likewise, if b_{2} ∈ {a_{1}, a_{2}}, type (II).

Suppose b_{2} = a_{3} . If b_{1} = a_{4} , type (I). So b_{1} ≠ a_{4} , 𝒜′_{3} = α_{1}, α_{2}, β_{2} , ℬ′_{2} = β_{1}, α_{3} , ℬ′′_{2} = β_{1}, ε_{t,h} and ℬ′_{3} = β_{1}, β_{2}, ε_{h,t} are all chains, Proposition 3.4(e), see the first to fourth wire-frames in Figure 15.

Suppose b_{2} = a_{4} . If b_{1} ∈ {a_{2}, a_{3}}, type (I) or (II). So b_{1} = a_{1} or b_{1} ∈ {a_{5}, a_{6}}.

For b_{1} = a_{1} we have that b_{4} = a_{2} . The arc set 𝒜_{3} ∪ ℬ_{3} divides 𝒪 into four components having 6, 4, 3 and 3 geodesic boundary pieces respectively. Label these by 𝒪_{6} , 𝒪_{4} , 𝒪_{3} and 𝒪′_{3} so that α_{2} lies on the boundary of 𝒪_{3} . Let c denote the cone point lying off 𝒜_{3} ∪ ℬ_{3} . We note that both 𝒜′_{2} = α_{1}, β_{1} , ℬ′_{2} = β_{1}, α_{1} are chains and so α_{2}, β_{2} are both shortest arcs.

Suppose c ∈ 𝒪_{4} , see the fifth wire-frame. Let α′_{1} (respectively β′_{1} ) denote the arc between c and a_{2} (respectively b_{2} ) in 𝒪_{4} . Both 𝒜′_{3} = α_{1}, α_{2}, ε_{t,t}, ℬ′_{3} = β_{1}, β_{2}, ε_{t,t} are chains and so max{l(α_{3}), l(β_{3})} ≤ l(ε_{t,t}). By Proposition 3.5(v), l(α′_{1}) < l(α_{1}) or l(β′_{1}) < l(β_{1}).

Now suppose c ∈ 𝒪_{6} ; see the sixth wire-frame. Let α′_{3} (respectively β′_{3} ) denote the arc between c and a_{3} (respectively b_{3} ) in 𝒪_{6} . Again both 𝒜′_{3} = α_{1}, α_{2}, ε_{t,t}, ℬ′_{3} = β_{1}, β_{2}, ε_{t,t} are chains and so max{l(α_{3}), l(β_{3})} ≤ l(ε_{t,t}). Since α_{2}, β_{2} are both shortest arcs l(α_{2}) = l(β_{2}). We can now apply Proposition 3.5(vii): l(α′_{3}) < l(α_{3}) or l(β′_{3}) < l(β_{3}). As 𝒜′′_{3} = α_{1}, α_{2}, α′_{3} , ℬ′′_{3} = β_{1}, β_{2}, β′_{3} are both chains we have a contradiction.

For c ∈ 𝒪_{3} (equivalently c ∈ 𝒪′_{3} ) we can apply Theorems 2.12 and 3.6; see Figure 20.

For b_{1} ∈ {a_{5}, a_{6}} we first suppose b_{4} = a_{1} . The arc set Υ = α_{1}∪α_{2}∪α_{3}∪β_{2}∪ β_{3} divides 𝒪 into three components, having 7, 4 and 3 geodesic boundary pieces respectively. Label these components 𝒪_{7} , 𝒪_{4} and 𝒪_{3} respectively. Either b_{1} ∈ 𝒪_{7} , see the seventh wire-frame, or b_{1} ∈ 𝒪_{3} , see the eighth wire-frame. For b_{1} ∈ 𝒪_{7} (respectively b_{1} ∈ 𝒪_{3} ) label the arc between b_{1}, b_{3} in 𝒪_{7} (respectively 𝒪_{3} ) by β′_{1} . Since 𝒜′_{3} = α_{1}, α_{2}, ε_{t,t}, ℬ′_{3} = β_{1}, β_{2}, ε_{t,t} are both chains max{l(α_{3}), l(β_{3})} ≤ l(ε_{t,t}). By Proposition 3.5(iii), l(α_{2}) < l(α_{1}) or l(β′_{1}) < l(β_{1}) contradicting α_{1}, β_{1} both being shortest arcs.

For this last example the same argument was adapted to the two different topological configurations. This also holds for all the remaining examples. Accordingly in Figure 15 we have pictured only one topological configuration to illustrate the argument used.

Figure 15. Applications of Proposition 3.5 for i_{1} = 3, j_{1} = 3

If b_{4} = a_{2} we can again apply Proposition 3.5(iii); see the ninth wire-frame. So b_{2} ∈ {a_{5}, a_{6}}. If b_{1} = a_{1} then b_{4} = a_{2} ; see the tenth wire-frame. The arc set 𝒜_{3} ∪ ℬ_{3}

divides 𝒪 into three components having 6, 6 and 3 geodesic boundary pieces respectively. Label these components 𝒪_{6} , 𝒪′_{6} and 𝒪_{3} respectively so that α_{2} lies on the boundary of 𝒪_{6} . In 𝒪_{6} , label by α′_{1} the arc between b_{1}, a_{3} ; by α′′_{1} the arc between b_{2}, a_{2} ; and by α′′′_{1} the arc between b_{2}, a_{3} . Again 𝒜′_{3} = α_{1}, α_{2}, ε_{t,t}, ℬ′_{3} = β_{1}, β_{2}, ε_{t,t} are both chains max{l(α_{3}), l(β_{3})} ≤ l(ε_{t,t}). Since β_{1} , the arc disjoint from α_{3} ∪β_{3} , is a shortest arc, l(β_{1}) ≤ l(ε_{h,t}) and we can apply Proposition 3.5(i): max{l(α′_{1}), l(α′′_{1}), l(α′′′_{1})} < l(α_{1}) contradicting α_{1} being a shortest arc.

If b_{1} = a_{2} then b_{4} = a_{1} and we can again apply Proposition 3.5(i); see the eleventh wire-frame. For b_{1} = a_{3} we can apply Proposition 3.4(b); see the 12th and 13th wire-frame. So b_{1} = a_{4} and we can apply Proposition 3.5(iii); see the 14th and 15th wire-frames.

Now j_{1} = 4. First we suppose b_{5} ∈ {a_{1}, a_{2}}. If b_{4} ∈ {a_{1}, a_{2}}, type (I). So b_{4} ∈ {a_{5}, a_{6}}. If b_{3} ∈ {a_{1}, a_{2}}, type (II).

Suppose b_{3} = a_{3} . If {b_{1}, b_{2}} ∋ a_{4} , type (I) or (II). If {b_{1}, b_{2}} ∌ a_{4} we can apply Proposition 3.4(e); see the first and second wire-frames, Figure 16, where we have pictured the possibilities for b_{5} = a_{1} .

Suppose b_{3} = a_{4} . If {b_{1}, b_{2}} ∋ a_{3} or b_{2} = a_{2} , type (I) or type (II). So b_{2} = a_{1} or b_{2} ∈ {a_{5}, a_{6}}.

If b_{2} = a_{1} then b_{5} = a_{2} , b_{1} ∈ {a_{5}, a_{6}} and we can apply Proposition 3.5(i); see the third wire-frame.

For b_{2} ∈ {a_{5}, a_{6}} either b_{1} = a_{1} , b_{5} = a_{2} or b_{1} = a_{2} , b_{5} = a_{1} and we can apply Proposition 3.5(i); see the fourth and fifth wire-frames.

So b_{3} ∈ {a_{5}, a_{6}}.

Suppose b_{2} = a_{1} and so b_{5} = a_{2} . Either b_{1} = a_{3} or b_{1} = a_{4} .
For b_{1} = a_{3} , {b_{1}, b_{2}} = {a_{1}, a_{3}} and we can apply Proposition 2.10.

For b_{1} = a_{4} we note that ℬ′_{2} = β_{1}, α_{1} is a chain and so β_{2} is a shortest arc. We can apply Proposition 3.5(i); see the sixth wire-frame.

Suppose b_{2} = a_{2} and so b_{5} = a_{1} . Either b_{1} = a_{3} or b_{1} = a_{4} .

For b_{1} = a_{3} , see the seventh wire-frame. The arc set 𝒜_{3} ∪ ℬ_{4} divides 𝒪 into three components having 6, 5 and 4 geodesic boundary pieces respectively. Label these components 𝒪_{6} , 𝒪_{5} and 𝒪_{4} respectively. In 𝒪_{6} , label by α′_{1} (respectively α′′_{1} ) the arc between b_{2}, a_{4} (respectively b_{3}, a_{1} ). We have that 𝒜′_{3} = α_{1}, α_{2}, ε_{t,t}, ℬ′_{4} = β_{1}, β_{2}, β_{3}, ε_{h,t} are both chains l(α_{3}) ≤ l(ε_{t,t}), l(β_{3}) ≤ l(ε_{h,t}). Now ℬ′_{2} = β_{1}, α_{1} is a chain and so β_{2} —the arc disjoint from α_{3} ∪ β_{4} —is a shortest arc, l(β_{2}) ≤ min{l(ε_{t,t}), l(ε_{t,h})} and so we can apply Proposition 3.5(ii): max{l(α′_{1}), l(α′′_{1})} < l(α_{1}) contradicting α_{1} being a shortest arc.

For b_{1} = a_{4} , type (II).

Suppose b_{2} = a_{3} . If b_{1} = a_{1} , we can apply Proposition 2.10. If b_{1} = a_{2} then b_{5} = a_{1} and we can apply Proposition 3.4(e); see the eighth wire-frame. If b_{1} = a_{4} , type (I).

So b_{2} = a_{4} .

If b_{1} = a_{1} then b_{5} = a_{2} and we note that ℬ′_{2} = β_{1}, α_{1} is a chain and so β_{2} is a shortest arc. We can apply Proposition 3.5(iii) to contradict α_{1}, β_{2} both being shortest arcs; see the ninth wire-frame.

So b_{1} ∈ {a_{2}, a_{3}}, type (I) or (II). Therefore: {a_{1}, a_{2}} ∌ b_{5} and a_{4} ∈ {b_{1}, b_{2}, b_{3}}.

If a_{3} ∈ {b_{1}, b_{2}, b_{3}}, type (I) or (II). So a_{3} = b_{6} .

Figure 16. Applications of Propositions 3.4 and 3.5 for i_{1} = 3, j_{1} = 4

Suppose a_{2} = b_{1} . If a_{4} = b_{2} , type (II). If a_{4} = b_{3} then a_{1} = b_{2} or a_{1} = b_{4} . We can apply Proposition 3.5(i) or (ii) respectively; see the tenth and eleventh wire-frames.

If a_{2} = b_{2} , type (II).

Suppose a_{2} = b_{3} . If a_{4} = b_{2} , type (I). If a_{4} = b_{1} then a_{1} = b_{2} or a_{1} = b_{4} . For a_{1} = b_{2} we can apply Proposition 3.5(i); see the 12th wire-frame.

For a_{1} = b_{4} , see the 13th wire-frame. The arc set 𝒜_{3}∪ℬ_{4} divides 𝒪 into three components having 6, 5 and 4 geodesic boundary pieces respectively. Label these components 𝒪_{6} , 𝒪_{5} and 𝒪_{4} respectively. In 𝒪_{6} , label by β′_{1} (respectively β′′_{1} ) the arc between b_{2}, b_{5} (respectively b_{3}, b_{1} ). We have that 𝒜′_{3} = α_{1}, α_{2}, ε_{t,h}, ℬ′_{4} = β_{1}, β_{2}, β_{3}, ε_{t,t} are both chains and so l(α_{3}) ≤ l(ε_{t,h}), l(β_{4}) ≤ l(ε_{t,t}). We know that ℬ′_{2} = β_{1}, ε_{h,h} is a chain and so l(β_{2}) ≤ l(ε_{h,h}). If l(ε_{h,h}) ≤ l(ε_{t,t}) then l(β_{2}) ≤ min{l(ε_{h,h}), l(ε_{t,t})} and we can apply Proposition 3.5(ii): max{l(β′_{1}), l(β′′_{1})} < l(β_{1}) contradicting β_{1} being a shortest arc. So suppose l(ε_{t,t}) ≤ l(ε_{h,h}). We can now apply Proposition 3.5(iv): l(β′_{1}) < l(β_{1}) or l(α_{2}) < l(α_{1}) contradicting β_{1}, α_{1} both being shortest arcs.

So a_{2} = b_{4} and we can apply Proposition 3.4(e), see the 14th and 15th wireframe, where we pictured the possibilities for a_{4} = b_{1} .

Finally we consider i_{1} = 4, j_{1} = 4. Up to relabelling, we may suppose that b_{5} ∈ {a_{1}, a_{2}, a_{3}}. If b_{4} ∈ {a_{1}, a_{2}, a_{3}}, type (I) or (II). So b_{4} = a_{6} .

Suppose b_{3} ∈ {a_{1}, a_{2}, a_{3}}. Either {b_{5}, b_{3}} ∋ a_{2} , type (II) or {b_{5}, b_{3}} = {a_{1}, a_{3}}.

First: (b_{5}, b_{3}) = (a_{1}, a_{3}). If b_{2} = a_{2} either b_{1} = a_{4} or b_{1} = a_{5} and we can apply Proposition 3.5(ii) or (i) respectively; see the first and second wire-frames, Figure 17. If b_{2} ∈ {a_{4}, a_{5}} either b_{1} = a_{2} and we can apply Proposition 2.10 or b_{1} ∈ {a_{4}, a_{5}}, type (I).

Next: (b_{5}, b_{3}) = (a_{3}, a_{1}). If b_{2} = a_{2} either b_{1} = a_{4} or b_{1} = a_{5} and we can apply Proposition 3.5(ii) or (i) respectively, see the third and fourth wire-frames. If b_{2} ∈ {a_{4}, a_{5}} either b_{1} = a_{2} and we can apply Proposition 2.10, or b_{1} ∈ {a_{4}, a_{5}}, type (I).

Suppose b_{3} = a_{4} . For {b_{1}, b_{2}} ∌ a_{5} we can apply Proposition 3.4(e); see the fifth and sixth wire-frames, where we have pictured the possibilities for b_{5} = a_{1} . For {b_{1}, b_{2}} ∋ a_{5} , type (I) or (II). So b_{3} = a_{5} .

Figure 17. Applications of Propositions 3.4 and 3.5 for i_{1} = 4, j_{1} = 4

Suppose b_{2} = a_{1} . If b_{1} = a_{2} we can apply Proposition 3.5(i); see the sixth wire-frame. If b_{1} = a_{3} we can apply Proposition 2.10. If b_{1} = a_{4} , type (II).

Suppose b_{2} = a_{2} . If b_{1} = a_{1} or b_{1} = a_{3} and we can apply Proposition 3.5(i); see the eighth and ninth wire-frames. If b_{1} = a_{4} , type (II).

So b_{2} ∈ {a_{3}, a_{4}}, type (I) or (II).

Let Z be an arc set that contains no crossing arcs and has the combinatorial pattern of the edge set of a triangular prism. Label cone points so that there is a Z arc between c_{n}, c_{n+1} and between c_{n}, c_{n+2} and between c_{n+1}, c_{n+3} for n = 0, 2, 4. Subscript addition is modulo 6. Label the Z arc between c_{n}, c_{n+1} by ζ_{n,n+1} , et cetera. Label by ζ_{n,n+3} (respectively ζ_{n+1,n+2} ) the arc between c_{n}, c_{n+3} (respectively c_{n+1}, c_{n+2} ) that does not cross any arc in Z for n = 0, 2, 4. In Figure 18 we have pictured an orbifold as triangular prism wire-frame with some of these arcs drawn in thick black.

Proposition 3.5. We have the following:

(i) If max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}) and l(ζ_{0,1}) ≤ l(ζ_{4,5}) then min{l(ζ_{0,3}), l(ζ_{1,3}), l(ζ_{1,2})} < l(ζ_{0,2}).

(ii) If l(ζ_{2,5}) ≤ l(ζ_{3,5}), l(ζ_{3,4}) ≤ l(ζ_{4,5}) and l(ζ_{0,1}) ≤ min{l(ζ_{2,3}), l(ζ_{4,5})} then min{l(ζ_{0,3}), l(ζ_{1,2})} < l(ζ_{0,2}).

(iii) If max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}) then either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}).

(iv) If l(ζ_{2,5}) ≤ l(ζ_{3,5}) and l(ζ_{3,4}) ≤ l(ζ_{4,5}) ≤ l(ζ_{2,3}) then either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}).

(v) If max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}) then either l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}).

(vi) If l(ζ_{2,5}) ≤ l(ζ_{3,5}) and l(ζ_{3,4}) ≤ l(ζ_{4,5}) then either l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}).

(vii) If max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}), l(ζ_{2,3}) = l(ζ_{4,5}) and ζ_{0,2}, ζ_{4,0} are shortest arcs, then l(ζ_{1,3}) < l(ζ_{3,4}) or l(ζ_{5,1}) < l(ζ_{2,5}).

Proof. Choose the distinguished disjoint triple of arcs Γ = ζ_{0,1} ∪ ζ_{2,3} ∪ ζ_{4,5} . We label the common perpendiculars to the pair of pants 𝒪 \ Γ by p_{0,2}, p_{2,4}, p_{4,0} .

Cutting 𝒪 open along Γ, ζ_{1,3}, ζ_{5,1} we obtain a simply connected domain Ω. Choose a lift of Ω in the universal cover of the pair of pants 𝒪 \ Γ. Without confusion, we shall use the same labels for geodesics having non-trivial intersection with Ω as on 𝒪. We label the orbits of cone points on the boundary of Ω in cyclic order: c_{0} , c_{1} , c_{3} , c_{2} , c′_{3} , c′_{1} , c′_{5} , c_{4} , c′′_{5} , c′′_{1} , so that ∗_{i} ∈ c̃_{i}. The points c_{0} , c_{2} , c_{4} lie at the midpoints of edges, the other points lie at vertices. Label the component of ζ̃_{0,1} containing c′_{1} by ζ′_{0,1} ; see Figure 18.

Label the common perpendiculars to ζ_{0,1}, ζ_{2,3}, ζ_{4,5} by p_{0,2}, p_{2,4}, p_{4,0} . Let ℋ denote the right hexagon bounded by p_{2,4} ∪ ζ_{2,3} ∪ p_{0,2} ∪ ζ_{0,1} ∪ p_{0,4} ∪ ζ_{4,5} . We shall refer to the inside and outside of p_{0,2} so that ℋ lies inside p_{0,2} . Similarly for p_{2,4}, p_{4,0} .

(i) We suppose that max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}), l(ζ_{0,1}), l(ζ_{0,1}) ≤ l(ζ_{4,5}) and l(ζ_{0,3}) ≥ l(ζ_{0,2}) and show that l(ζ_{1,2}) ≤ l(ζ_{0,2}). Unless all the inequalities are equalities this inequality will be strict. If all the inequalities are equalities then we show that l(ζ_{1,3}) < l(ζ_{0,2}).

Consider 𝒬′_{2,3} bounded by ζ_{4,5} ∪ p_{2,4} ∪ ζ_{2,3}∪ ⊥ c_{2}c′_{3} and 𝒬′_{5,4} bounded by ζ_{2,3} ∪ p_{2,4} ∪ ζ_{4,5}∪ ⊥ c′_{5}c_{4} , a pair of right quadrilaterals. Recall that ⊥ XY denotes the bisector of a disjoint pair of points X, Y ∈ H^{2} . Any right quadrilateral 𝒬 bounded by A ∪ B ∪ C ∪ D, will be such that the A edge and D edge meet in an acute angle (𝒬 is ‘finite’ or ‘a trirectangle’) or the A edge and D edge do not meet (𝒬 is ‘infinite’).

Claim 1. Both 𝒬′_{2,3} , 𝒬′_{5,4} are strictly outside p_{2,4} , c′_{5} is on the ζ_{4,5} edge of 𝒬′_{2,3} and c′_{3} is on the ζ_{2,3} edge of 𝒬′_{5,4} .

If either c′_{3} or c′_{5} is inside p_{2,4} then either ^{}c̸_{2}c′_{3}c′_{5} ≥ π/2 or ^{}c̸′_{3}c′_{5}c_{4} ≥ π/2 and so either l(ζ_{3,5}) < l(ζ_{2,5}) or l(ζ_{3,5}) < l(ζ_{3,4}). So both c′_{3} , c′_{5} are strictly outside p_{2,4} . As above, we use ^{}x̸yz to denote the angle at the y vertex of the triangle spanned by xyz.

Likewise if either ⊥ c_{2}c′_{3} or ⊥ c′_{5}c_{4} is inside p_{2,4} then either l(ζ_{3,5}) < l(ζ_{2,5}) or l(ζ_{3,5}) < l(ζ_{3,4}), since c′_{3} , c′_{5} are strictly outside p_{2,4} . So ⊥ c_{2}c′_{3} , ⊥ c′_{5}c_{4} and hence 𝒬′_{2,3} , 𝒬′_{5,4} are both strictly outside p_{2,4} .

If 𝒬′_{2,3} is finite and c′_{5} is strictly beyond its acute vertex then l(ζ_{3,5}) < l(ζ_{2,5}). Similarly if 𝒬′_{5,4} is finite and c′_{3} is strictly beyond its acute vertex then l(ζ_{3,5}) < l(ζ_{3,4}). This completes the claim.

Figure 18. Arcs and lift for (i) and (ii)

Consider a third quadrilateral 𝒬_{3,2} bounded by ζ_{0,1} ∪ p_{0,2} ∪ ζ_{2,3}∪ ⊥ c_{3}c_{2} . We compare edge-lengths of 𝒬_{3,2} , 𝒬′_{2,3} . Since d(c_{3}, c_{2}) = d(c_{2}, c′_{3}) = l(ζ_{2,3}) = d(p_{0,2}, p_{2,4}), the ζ_{2,3} edges of 𝒬_{3,2}𝒬′_{2,3} are the same length. Also 𝒬_{3,2} is strictly inside p_{0,2} since 𝒬′_{2,3} is strictly outside p_{2,4} . Now l(ζ_{0,1}) ≤ l(ζ_{4,5}) and so, by the geometry of right hexagons l(p_{0,2}) ≥ l(p_{2,4}). That is, the p_{0,2} edge of 𝒬_{3,2} is longer than the p_{2,4} edge of 𝒬′_{2,3} and so, by the geometry of right quadrilaterals, the ζ_{0,1} edge of 𝒬_{3,2} is longer than the ζ_{4,5} edge of 𝒬′_{2,3} .

If c_{0} is outside p_{0,2} or on the interior of the ζ_{0,1} edge of 𝒬_{3,2} , l(ζ_{0,3}) < l(ζ_{0,2}), contradicting our supposition. So 𝒬_{3,2} is finite and c_{0} beyond its ζ_{0,1} edge which is longer than the ζ_{4,5} edge of 𝒬′_{2,3} which contains c′_{5} . So d(c_{0}, p_{0,2}) ≥ d(c′_{5}, p_{2,4}). Since ⊥ c′_{5}c_{4} is strictly outside p_{2,4} and d(c_{0}, c_{1}) = l(ζ_{0,1}) ≤ l(ζ_{4,5}) = d(c′_{5}c_{4}), ⊥ c_{0}c_{1} is strictly inside p_{0,2} and d(⊥ c_{0}c_{1}, p_{0,2}) ≥ d(⊥ c′_{5}c_{4}, p_{2,4}).

Consider a fourth quadrilateral 𝒬_{0,1} bounded by ζ_{2,3} ∪ p_{0,2} ∪ ζ_{0,1}∪ ⊥ c_{0}c_{1} . We compare edge lengths of 𝒬_{0,1} , 𝒬′_{5,4} . We have shown that 𝒬_{0,1} is strictly inside p_{0,2} and that the ζ_{0,1} edge of 𝒬_{0,1} is longer than the ζ_{4,5} edge of 𝒬′_{5,4} . Since l(p_{0,2}) ≥ l(p_{2,4}), the ζ_{2,3} edge of 𝒬_{0,1} is longer than the ζ_{2,3} edge of 𝒬′_{5,4} . As c′_{3} is on the ζ_{2,3} edge of 𝒬′_{5,4} and d(c′_{3}, p_{2,4}) = d(c_{2}, p_{0,2}), c_{2} is on the ζ_{2,3} edge of 𝒬_{0,1} . So l(ζ_{1,2}) ≤ l(ζ_{0,2}) as required.

Finally, l(ζ_{3,4}) = l(ζ_{2,5}) = l(ζ_{3,5}), l(ζ_{0,1}) = l(ζ_{4,5}), l(ζ_{0,3}) = l(ζ_{0,2}) implies l(ζ_{1,3}) < l(ζ_{0,2}). So 𝒬_{0,1} is finite and c_{2} is at its acute vertex. Now c_{3} is either outside p_{0,2} or on the interior of the ζ_{2,3} edge of 𝒬_{0,1} : l(ζ_{1,3}) < l(ζ_{0,3}) = l(ζ_{0,2}).

(ii) We suppose l(ζ_{2,5}) ≤ l(ζ_{3,5}), l(ζ_{3,4}) ≤ l(ζ_{4,5}), l(ζ_{0,1}) ≤ min{l(ζ_{2,3}), l(ζ_{4,5})} and l(ζ_{0,3}) ≥ l(ζ_{0,2}) and show that l(ζ_{1,2}) < l(ζ_{0,2}).

Claim 2. 𝒬′_{2,3} , c′_{5} are both strictly outside p_{2,4} , c′_{5} is on the ζ_{4,5} edge of 𝒬′_{2,3} and d(c′_{5}, p_{2,4}) > d(c′_{3}, p_{2,4}).

Suppose that 𝒬′_{2,3} is strictly inside p_{2,4} .

If 𝒬′_{2,3} is finite and c′_{5} is beyond the acute vertex of 𝒬′_{2,3} then ^{}c̸′_{3}c′_{5}c_{4} > π/2, l(ζ_{4,5}) < l(ζ_{3,4}). If c′_{5} is on the interior of the ζ_{4,5} edge of 𝒬′_{2,3} or outside p_{2,4} , l(ζ_{3,5}) < l(ζ_{2,5}). Similar arguments hold for 𝒬′_{2,3} trivial or infinite.

Suppose 𝒬′_{2,3} (and hence c′_{3} ) is strictly outside p_{2,4} .

If c′_{5} is inside p_{2,4} we again have ^{}c̸′_{3}c′_{5}c_{4} > π/2, l(ζ_{4,5}) < l(ζ_{3,4}). If c′_{5} is outside p_{2,4} such that d(c′_{5}, p_{2,4}) ≤ d(c′_{3}, p_{2,4}) then ^{}c̸′_{3}c′_{5}c_{4} ≥ ^{}c̸_{2}c′_{3}c′_{5} . By inspection ^{}c̸_{2}c′_{3}c′_{5} > ^{}c̸_{4}c′_{3}c′_{5} so ^{}c̸′_{3}c′_{5}c_{4} > ^{}c̸_{4}c′_{3}c′_{5} , l(ζ_{4,5}) < l(ζ_{3,4}). If 𝒬′_{2,3} is finite and c′_{5} is strictly beyond its acute vertex, l(ζ_{3,5}) < l(ζ_{2,5}). The claim follows.

Recall that 𝒬_{3,2} is bounded by ζ_{0,1} ∪ p_{0,2} ∪ ζ_{2,3}∪ ⊥ c_{3}c_{2} . As 𝒬′_{2,3} is strictly outside p_{2,4} and the ζ_{2,3} edges of 𝒬_{3,2} , 𝒬′_{2,3} are the same length: 𝒬_{3,2} is strictly inside p_{0,2} . Again l(ζ_{0,1}) ≤ l(ζ_{4,5}) and so l(p_{0,2}) ≥ l(p_{2,4}). So the ζ_{0,1} edge of 𝒬_{3,2} is longer than the ζ_{4,5} edge of 𝒬′_{2,3} .

Recall that 𝒬_{0,1} is bounded by ζ_{2,3} ∪ p_{0,2} ∪ ζ_{0,1}∪ ⊥ c_{0}c_{1} . We show that 𝒬_{0,1} is strictly inside p_{0,2} . As with part (i), so that l(ζ_{0,3}) ≥ l(ζ_{0,2}): 𝒬_{3,2} is finite and c_{0} is beyond its acute vertex. So d(c_{0}, p_{0,2}) is longer than the ζ_{0,1} edge of 𝒬_{3,2} which is longer than the ζ_{4,5} edge of 𝒬′_{2,3} which is longer than d(c′_{5}, p_{2,4}) > d(c′_{3}, p_{2,4}) = d(c_{2}, p_{0,2}). That is d(c_{0}, p_{0,2}) > d(c_{2}, p_{0,2}). Since ⊥ c_{3}c_{2} is strictly inside p_{0,2} and d(c_{0}, c_{1}) = l(ζ_{0,1}) ≤ l(ζ_{2,3}) = d(c_{3}, c_{2}), ⊥ c_{0}c_{1} is strictly inside p_{0,2} and d(⊥ c_{0}c_{1}, p_{0,2}) > d(⊥ c_{3}c_{2}, p_{0,2}).

So 𝒬_{3,2} , 𝒬_{0,1} are both inside p_{0,2} , the ζ_{0,1} edge of 𝒬_{0,1} is strictly longer than the ζ_{2,3} edge of 𝒬_{3,2} . As 𝒬_{3,2} , 𝒬_{0,1} have p_{0,2} in common, the ζ_{2,3} edge of 𝒬_{0,1} is strictly longer than the ζ_{0,1} edge of 𝒬_{3,2} which is longer than the ζ_{4,5} edge of 𝒬′_{2,3} which is strictly longer than d(c′_{5}, p_{2,4}) > d(c′_{3}, p_{2,4}) = d(c_{2}, p_{0,2}). So c_{2} is on the interior of the ζ_{2,3} edge of 𝒬_{0,1} , l(ζ_{1,2}) < l(ζ_{0,2}).

(iii) We suppose that max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}) and show that either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}). From Claim 1, 𝒬′_{2,3} , 𝒬′_{5,4} are both strictly outside p_{2,4} and so 𝒬_{3,2} , 𝒬′′_{4,5} are strictly inside p_{0,2} , p_{4,0} respectively. Here 𝒬′′_{4,5} denotes the right quadrilateral bounded by ζ_{0,1} ∪ p_{4,0} ∪ ζ_{4,5} ⊥ c_{4}c′′_{5} .

Consider c_{0} , c′′_{1} on ζ_{0,1} . Either c_{0} is outside p_{0,2} or c′′_{1} is outside p_{4,0} . So c_{0} is strictly closer to c_{3} than to c_{2} or c′′_{1} is strictly closer to c′′_{5} than to c_{4} . That is either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}).

(iv) We suppose that l(ζ_{2,5}) ≤ l(ζ_{3,5}) and l(ζ_{3,4}) ≤ l(ζ_{4,5}) ≤ l(ζ_{2,3}) and show that either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}).

From Claim 2, 𝒬′_{2,3} , c′_{5} are both strictly outside p_{2,4} , d(c′_{5}, p_{2,4}) > d(c′_{3}, p_{2,4}). Since d(c′_{5}, c_{4}) = l(ζ_{4,5}) ≤ l(ζ_{2,3}) = d(c_{2}, c′_{3}), 𝒬′_{5,4} is strictly outside p_{2,4} . As with part (iii), we can conclude that either l(ζ_{0,3}) < l(ζ_{0,2}) or l(ζ_{5,1}) < l(ζ_{4,1}).

(v) We suppose that max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}) and show that either l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}). From Claim l, c′_{3} , c′_{5} are both strictly outside p_{2,4} and so c_{2} and c_{4} are strictly inside p_{0,2} and p_{4,0} respectively.

Consider ⊥ c_{0}c_{1} , ⊥ c′′_{1}c_{0} . Either ⊥ c_{0}c_{1} is outside p_{0,2} or ⊥ c′′_{1}c_{0} is outside p_{4,0} . So either c_{2} is strictly closer to c_{0} than to c_{1} or c_{4} is strictly closer to c_{0} than to c′′_{1} . That is either l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}).

(vi) We suppose that l(ζ_{2,5}) ≤ l(ζ_{3,5}), l(ζ_{3,4}) ≤ l(ζ_{4,5}) and show that l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}). From Claim 2, c′_{3} and c′_{5} are both strictly outside p_{2,4} . As with part (v) we can conclude that either l(ζ_{0,2}) < l(ζ_{1,2}) or l(ζ_{4,0}) < l(ζ_{4,1}).

(vii) We suppose that max{l(ζ_{2,5}), l(ζ_{3,4})} ≤ l(ζ_{3,5}), l(ζ_{2,3}) = l(ζ_{4,5}) and that ζ_{0,2} , ζ_{4,0} are both shortest arcs and show that l(ζ_{1,3}) < l(ζ_{3,4}) or l(ζ_{5,1}) < l(ζ_{2,5}).

By hypothesis l(ζ_{2,3}) = l(ζ_{4,5}) and so 𝒪 \ Γ has rotational symmetry R exchanging boundary components ζ_{2,3} , ζ_{4,5} and fixing ζ_{0,1} , p_{2,4} setwise. Gluing along ζ_{0,1} to recover 𝒪 \ ζ_{2,3} ∪ ζ_{4,5} , this symmetry is respected, exchanging cone points c_{0} , c_{1} . In the universal cover, the rotational symmetry of 𝒪 \ζ_{2,3} ∪ζ_{4,5} lifts to a rotational symmetry R such that R(ζ_{2,3}) = ζ_{4,5} , R(ζ_{0,1}) = ζ′_{0,1} , R(c_{0}) = c′_{1} and R(p_{2,4}) = p_{2,4} .

By Claim 1, ⊥ c_{2}c′_{3} and ⊥ c′_{5}c_{4} are both strictly outside p_{2,4} .

Figure 19. Arcs and lift for (vii)

Either (a) c_{2} , c_{4} are both strictly inside p_{2,4} or (b) one of c_{2} , c_{4} is outside p_{2,4} . We shall give the argument for (b); a similar argument holds for (a). We suppose c_{2} is outside p_{2,4} and show that l(ζ_{1,3}) < l(ζ_{3,4}). (If c_{4} is outside p_{2,4} , the same argument shows that l(ζ_{5,1}) < l(ζ_{2,5}).)

Claim 3. c_{4} is inside p_{2,4} and d(c_{2}, p_{2,4}) ≤ d(c_{4}, p_{2,4}) < l(ζ_{4,5})/2.

Consider the bisector ⊥ c_{0}c′_{1} . By rotational symmetry ⊥ c_{0}c′_{1} passes through the midpoint of p_{2,4} . As ζ_{0,2} is a shortest arc, c_{2} ∈ ζ_{2,3} is to the c_{0} side of ⊥ c_{0}c′_{1} . So either ζ_{2,3} is strictly to the c_{0} side of ⊥ c_{0}c′_{1} or ⊥ c_{0}c′_{1} intersects ζ_{2,3} outside p_{2,4} , beyond c_{2} .

By rotational symmetry, the former corresponds to ζ_{4,5} ∋ c_{4} lying strictly to the c′_{1} side of ⊥ c_{0}c′_{1} . This contradicts ζ_{4,0} being a shortest arc. So ⊥ c_{0}c′_{1} intersects ζ_{2,3} (respectively ζ_{4,5} ) outside (inside) p_{2,4} , beyond c_{2} (between p_{2,4} , c_{4} ). As ⊥ c′_{5}c_{4} is strictly outside p_{2,4} , d(c_{4}, p_{2,4}) < d(c′_{5}, p_{2,4}) and so d(c_{4}, p_{2,4}) < l(ζ_{4,5})/2. This completes the claim.

By rotational symmetry R(c_{4} ) is outside p_{2,4} and d(c_{2}, p_{2,4}) ≤ d(R(c_{4}), p_{2,4}) < l(ζ_{2,3})/2.

Claim 4. d(R(c_{4}), c′_{1}) ≤ d(R(c_{4}), c_{4}).

We show that d(c_{4}, c_{0}) ≤ d(c_{4}, R(c_{4})), which by rotational symmetry is equivalent. By hypothesis ζ_{4,0} is a shortest arc and so d(c_{4}, c_{0}) = l(ζ_{4,0}) ≤ l(ζ_{2,4}) = d(c_{4}, c_{2}). By Claim 3 and by inspection d(c_{4}, c_{2}) ≤ d(c_{4}, R(c_{4})).

Now c′_{3} is outside p_{2,4} such that d(c′_{3}, p_{2,4}) = l(ζ_{2,3})+d(c_{2}, p_{2,4}) > l(ζ_{2,3})/2 > d(R(c_{4}), p_{2,4}). So by Claim 4 and by inspection, d(c′_{3}, c′_{1}) = l(ζ_{1,3}) < l(ζ_{3,4}) = d(c′_{3}, c_{4}).

We extend the arc set K ∪Λ by H = ∪η_{l} where η_{l} is between c_{1} , c_{3} crossing only κ_{0,l} ⊂ K.

Theorem 3.6. Either l(κ_{0,1}) < l(η_{4}) or l(κ_{0,5}) < l(κ_{3,5}) or l(λ_{2}) < l(λ_{0}).

Figure 20. Application, arcs and lift of Ω in Theorem 3.6

Consider 𝒜_{i_{1}}, ℬ_{j_{1}} with i_{1} = 3, j_{1} = 3 and b_{4} = a_{2} , b_{3} ∈ {a_{5}, a_{6}}, b_{2} = a_{4} , b_{1} = a_{1} such that c ∈ 𝒪_{3} ; see Figure 20. Since 𝒜′_{2} = α_{1}, β_{1} is a chain, α_{2} is a shortest arc. We set α_{1} = κ_{2,3} , α_{2} = κ_{3,5} , α_{3} = λ_{0} , β_{1} = κ_{2,4} , β_{2} = κ_{1,4} , β_{3} = η_{4} . This arc set extends uniquely to the arc set K ∪ Λ ∪ H. Both 𝒜′_{3} = α_{1}, α_{2}, λ_{1} , 𝒜′′_{3} = α_{1}, α_{2}, λ_{3} are chains, so l(λ_{0}) ≤ min{l(λ_{1}), l(λ_{3})}. Since α_{1} is a shortest arc, Theorem 2.12 implies min{l(λ_{1}), l(λ_{3})} ≤ l(λ_{2}). So l(λ_{0}) ≤ l(λ_{2}). By Theorem 3.6 either l(κ_{0,1}) < l(η_{4}) or l(κ_{0,5}) < l(κ_{3,5}) or l(λ_{2}) < l(λ_{0}). As ℬ′_{3} = β_{1}, β_{2}, κ_{0,1} is a chain and α_{2} = κ_{3,5} is a shortest arc, we have a contradiction.

Proof. We suppose l(η_{4}) ≤ l(κ_{0,1}), l(κ_{3,5}) ≤ l(κ_{0,5}) and show that l(λ_{2}) < l(λ_{0}). Cut 𝒪 open along κ_{1,5} ∪ κ_{1,4} ∪ λ_{2} and consider 𝒪′_{2} , the component not containing c_{2} . Cut 𝒪′_{2} open along κ_{3,0} so as to obtain an annulus. Making a further cut along κ_{3,5} we obtain a simply connected domain Ω. Choose a lift of Ω in the universal cover of the annulus 𝒪′_{2} \ κ_{3,0} .

Label the geodesics around the boundary of Ω by κ_{3,0} , κ_{3,5} , λ_{2} , κ_{1,4} , κ_{1,5} , κ′_{3,5} , in cyclic order. Give the other geodesics having non-trivial intersection with Ω the same labels as on 𝒪. In the same cyclic order, label orbits of cone points around the boundary of Ω by c_{0} , c_{3} , c_{5} , c_{4} , c_{1} , c′_{5} , c′_{3} . Let P_{5} , P_{4} , P_{1} , P′_{5} denote the perpendiculars to κ_{3,0} from c_{5} , c_{4} , c_{1} , c′_{5} respectively.

Now: l(η_{4}) ≤ l(κ_{0,1}) implies that P_{1} is closer to c_{3} than to c_{0} and l(κ_{3,5}) ≤ l(κ_{0,5}) implies that P′_{5} is closer to c′_{3} than to c_{0} . It follows that d(P_{1}, P′_{5}) ≥ l(κ_{3,0}).

The perpendiculars lie in the order P_{5}, P_{4}, P_{1}, P′_{5} and so d(P_{4}, P′_{5}) > d(P_{1}, P′_{5}) and hence d(P_{5}, P_{4}) < l(κ_{3,0}). So the κ_{3,0} edge of the birectangle 𝒬_{2,3} is strictly shorter than the κ_{3,0} edge of the birectangle 𝒬_{0,3} . Therefore l(λ_{2}) < l(λ_{0}).

Theorem 3.7. If l(λ_{0}) ≤ 1(λ_{2}), l(κ_{0,1}) ≤ min{l(κ_{0,l}), l(κ_{3,l}), l(η_{l})} and l(κ_{1,2}) = l(κ_{3,0}) ≤ min{l(κ_{1,l}), l(κ_{2,l}), l(κ_{3,l})}, for l = 4, 5, then 𝒪 is the octahedral orbifold.

Consider 𝒜_{i_{1}}, ℬ_{j_{1}} with i_{1} = 2, j_{1} = 3 and a_{3} ∈ {b_{1}, b_{2}}, {a_{2}, a_{1}} = {b_{5}, b_{6}}; see Figure 21. Suppose a_{3} = b_{1} . Set α_{1} = κ_{3,0} , α_{2} = κ_{0,1} , β_{1} = κ_{1,2} , β_{2} = κ_{2,5} , β_{3} = λ_{0} . This arc set extends uniquely to the arc set K ∪ Λ ∪ H. Each one of ℬ′_{3} = β_{1}, β_{2}, λ_{2} , 𝒜′_{2} = α_{1}, κ_{0,l}, 𝒜′′_{2} = α_{1}, κ_{3,l}, 𝒜′′′_{2} = α_{1}, η_{l} is a chain so l(λ_{0}) ≤ l(λ_{2}), l(κ_{0,1}) ≤ {l(κ_{0,l}), l(κ_{3,l}), l(η_{l})} for l = 4, 5. Also κ_{3,0} = α_{1} , κ_{1,2} = β_{1} are both shortest arcs, so we can apply Theorem 3.7: 𝒪 is the octahedral orbifold. This gives a contradiction since minimal chains on 𝒪ct lie in its set of shortest arcs, which contains no crossing arcs. Likewise for a_{3} = b_{2} .

Figure 21. Applications and arc set for Theorem 3.7

Proof. Let Γ = κ_{1,2} ∪ λ_{2} ∪ κ_{3,0} a distinguished disjoint triple of arcs. Let p_{3,1} denote the common perpendicular between κ_{3,0} , κ_{1,2} . The pair of pants 𝒪 \Γ has a rotational symmetry R exchanging κ_{1,2} , κ_{3,0} and fixing λ_{0} , λ_{2} , p_{3,1} setwise. Gluing along λ_{2} so as to recover 𝒪 \ κ_{1,2} ∪ κ_{3,0} this symmetry is respected. So R exchanges the birectangles 𝒬_{0,1} ↔ 𝒬_{0,3} , 𝒬_{2,1} ↔ 𝒬_{2,3} . That is, 𝒬_{0,1} , 𝒬_{0,3} (respectively 𝒬_{2,1} , 𝒬_{2,3} ) are isometric.

As we have observed many times above, since l(λ_{0}) ≤ l(λ_{2}) the κ_{3,0} edge of 𝒬_{0,3} is shorter than the κ_{3,0} edge of 𝒬_{2,3} . Here we observe that l(λ_{0}) ≤ l(λ_{2}) implies that ^{}c̸_{l}𝒬_{0,1} ≥ ^{}c̸_{l}𝒬_{2,1} . Summing angles we have that ^{}c̸_{4}𝒬_{0,3} + c̸_{5}𝒬_{0,3} ≥ ^{}c̸_{4}𝒬_{2,3} +^{}c̸_{5}𝒬_{2,3} .

The set κ_{1,2} ∪ P_{4,1} ∪ P_{4,3} ∪ κ_{3,0} ∪ P_{5,3} ∪ P_{5,1} divides 𝒪 into a pair of hexagons ℋ_{k} ⊃ λ_{k} for k = 0, 2. We have shown that both hexagons have rotational symmetry and that the κ_{1,2} , κ_{3,0} edges of ℋ_{0} are shorter than the κ_{1,2} , κ_{3,0} edges of ℋ_{2} . It follows that the κ_{1,2} , κ_{3,0} edges of ℋ_{0} are shorter than l(κ_{1,2}) = l(κ_{3,0}). Also ^{}c̸_{4}ℋ_{0} = ^{}c̸_{4}𝒬_{0,1} +^{}c̸_{4}𝒬_{0,3} ≥ ^{}c̸_{4}𝒬_{2,1} +^{}c̸_{4}𝒬_{2,3} = ^{}c̸_{4}ℋ_{2} . Since c̸_{4}ℋ_{0} +^{}c̸_{4}ℋ_{2} = π it follows that ^{}c̸_{4}ℋ_{0} = ^{}c̸_{5}ℋ_{0} ≥ π/2.

Note. The hypotheses are symmetric up to an exchange of labels: κ_{k,4} ↔ κ_{k,5} for k = 0, 1, 2, 3; c_{4} ↔ c_{5} ; and η_{4} ↔ η_{5} . So we may suppose that l(P_{4,1}) = l(P_{5,3}) ≥ l(P_{4,3}) = l(P_{5,1}).

Cut 𝒪 open along κ_{1,2} ∪ κ_{1,4} ∪ κ_{3,4} ∪ κ_{3,0} ∪ κ_{0,5} ∪ κ_{2,5} to obtain a pair of simply connected domains Ω_{0}, Ω_{2} containing λ_{0}, λ_{2} respectively. Take a lift of Ω_{0} to the universal cover of the pair of pants 𝒪 \ Γ. Without confusion use the same labels for geodesics having non-trivial intersection with Ω_{0} , for orbits of cone points lying on the boundary of Ω_{0} , for the common perpendicular between κ_{1,2}, κ_{3,0} , for the perpendiculars from c_{4}, c_{5} to κ_{1,2}, κ_{3,0} and for the rotational symmetric of hexagon ℋ_{0} bounded by κ_{1,2} ∪ P_{4,1} ∪ P_{4,3} ∪ κ_{3,0} ∪ P_{5,3} ∪ P_{5,1} . We may suppose that ℋ_{0} has a bottom, a top, a left and a right, so that κ_{1,2} is the bottom, κ_{3,0} is the top, c_{4} to the left and c_{5} is to the right. Label the unlabelled vertices of ℋ_{0} : h_{1} , h_{2} , h_{0} , h_{3} in anticlockwise order, beginning from bottom left.

The main part of the proof has two parts. The first part is to establish

Claim 1. c_{1} is strictly between P_{4,1} , P_{5,1} .

For c_{1} to the left of P_{4,1} we show that l(κ_{0,4}) < l(κ_{0,1}). For c_{1} right of P_{5,1} a similar argument shows that l(κ_{0,5}) < l(κ_{0,1}).

We use a different pants decomposition of 𝒪. Let Γ′ = κ_{3,0} ∪ κ_{1,4} ∪ κ_{2,5} . Label the common perpendiculars associated to this pants decomposition by p_{0,4} , p_{4,5} and p_{5,0} . Cut open along Γ′, κ_{1,2} , κ_{3,4} and κ_{0,5} so as to obtain two simply connected domains: Ω_{0} , Ω_{2} , as above. Again take a lift of Ω_{0} and use the same labels as before, with the additional labelling for the common perpendiculars to Γ′. So κ_{1,4} ∪ p_{4,5} ∪ κ_{2,5} ∪ p_{5,0} ∪ κ_{3,0} ∪ p_{0,4} bounds a right hexagon. Let p denote the common perpendicular to κ_{1,4}, p_{5,0} .

As c_{1} is to the left of P_{4,1} , ^{}c̸_{1}c_{4}h_{3} ≥ ^{}h̸_{1}c_{4}h_{3} = ^{}c̸_{4}ℋ_{0} ≥ π/2. So c_{4} is to the p_{4,5} side of p_{0,4} or equivalently, ⊥ c_{1}c_{4} is closer to p_{4,5} than to p_{0,4} . Now p is closer to p_{0,4} than to p_{4,5} , as l(κ_{3,0}) ≤ l(κ_{2,5}) and using the geometry of right hexagons. It follows that ⊥ c_{1}c_{4} lies to the p_{4,5} side of p and so ⊥ c_{1}c_{4} does not intersect κ_{3,0} and hence l(κ_{0,4}) < l(κ_{0,1}). This completes the claim.

Figure 22. Claims 1 and 2.

In the second part, we show that l(κ_{0,4}) ≤ l(κ_{0,1}) or l(κ_{0,5}) ≤ l(κ_{0,1}) with equality if and only if a certain set of conditions are satisfied. At the end we show that this set of conditions implies that 𝒪 is the octahedral orbifold.

The argument we use depends upon the positions of c_{0}, c_{1} : each part (i)–(viii) corresponds to a different configuration. However each part uses the following construction together with the angle comparison we establish in Claim 2.

For each part we define x to be a point on κ_{1,2} strictly between P_{4,1}, P_{5,1} . We define y to be the point on κ_{1,2} to the right of x such that d(x, y) = l(κ_{1,2}). Label angles so that φ = ^{}h̸_{1}c_{4}x, ψ = ^{}c̸_{4}xh_{1} , ψ′ = ^{}h̸_{2}xc_{5} , φ′ = ^{}x̸c_{5}h_{2} , φ′′ = ^{}h̸_{2}c_{5}y. Let m_{1,2} (respectively m_{3,0} ) denote the midpoint of h_{1}, h_{2} (respectively h_{3}, h_{0} ).

Claim 2. For (i), (ii), (iv), (v): ^{}R̸(x)c_{4}x ≥ π/2 − ψ and for (iii), (vi), (vii), (viii): ^{}R̸(x)c_{5}x ≥ π/2 − ψ′.

By rotational symmetry we have ^{}R̸(x)c_{4}x = ^{}R̸(x)c_{5}x. Below we show that R̸(x)c_{5}x ≥ π/2 − ψ′ for (i)–(viii). For (i), (ii), (iv), (v) the point x is to the left of m_{1,2} . So d(x, P_{4,1}) ≤ d(x, P_{5,1}). Since l(P_{4,1}) ≥ l(P_{5,1}), it follows that ψ ≥ ψ′. So the claim follows from the argument below.

Firstly: φ ≤ φ′′. Since d(x, y) = l(κ_{1,2}) ≥ d(P_{4,1}P_{5,1}) it follows that d(P_{4,1}, x) ≤ d(P_{5,1}, y). Also l(P_{4,1}) ≥ l(P_{5,1}).

Next: φ′ + φ′′ ≤ ψ′. For (i), (ii), (iii), (v)(a), (viii) (respectively (v)(b), (vi)) we have that y is to the right of c_{2} which is to the right of P_{5,1} (R(y) is to the left of c_{3} which is to the left of P_{4,3} ). It follows that d(y, c_{5}) ≥ d(c_{2}, c_{5}) = l(κ_{2,5}) ≥ l(κ_{1,2}) = d(y, x) (d(R(y), c_{4}) ≥ d(c_{3}, c_{4}) = l(κ_{3,4}) ≥ l(κ_{3,0}) = d(R(y), R(x))). This is equivalent to φ′ + φ′′ ≤ ψ′.

For (iv), (vii) we have that x is to the left of c_{1} which is to the left of P_{5,1} . It follows that d(x, c_{5}) ≥ d(c_{1}, c_{5}) = l(κ_{1,5}) ≥ l(κ_{1,2}) = d(x, y). Also x is to the right of m_{1,2} and so d(x, P_{5,1}) ≤ d(m_{1,2}, P_{5,1}) = d(P_{4,1}, P_{5,1})/2 ≤ l(κ_{1,2})/2. So d(y, P_{5,1}) ≥ d(x, P_{5,1}), d(y, c_{5}) ≥ d(x, c_{5}) ≥ d(x, y). Again, this is equivalent to φ′ + φ′′ ≤ ψ′.

So ^{}R̸(x)c_{4}x = ^{}c̸_{4}ℋ_{0}−(φ+φ′) ≥ π/2−(φ+φ′) ≥ π/2−(φ′′+φ′) ≥ π/2−ψ′ and we are done.

Note. It is an elementary exercise to show that m_{1,2} is to the right of p_{3,1} and that P_{0,1} (the perpendicular from c_{0} to κ_{1,2} ) is strictly between P_{4,1}, P_{5,1} .

First: c_{1} to the left of p_{3,1} .

(i): P_{0,1} to the left of c_{1} .

Let x = c_{1} . By inspection, ^{}c̸_{0}c_{4}x ≥ ^{}R̸(x)c_{4}x and by Claim 2, ^{}R̸(x)c_{4}x ≥ π/2 − ψ. Also by inspection, π/2 − ψ ≥ ^{}c̸_{0}xc_{4} . So ^{}c̸_{0}c_{4}c_{1} ≥ ^{}c̸_{0}c_{1}c_{4} or equivalently d(c_{0}, c_{4}) ≤ d(c_{0}, c_{1}) : l(κ_{0,4}) ≤ l(κ_{0,1}).

(ii): P_{0,1} between c_{1} , p_{3,1} .

Let x = P_{0,1} ∩ κ_{1,2} . By inspection, ^{}c̸_{0}c_{4}x ≥ ^{}R̸(x)c_{4}x and by Claim 2, R̸(x)c_{4}x ≥ π/2−ψ = ^{}c̸_{0}xc_{4} . So d(c_{0}, c_{4}) ≤ d(c_{0}, x). Now x is the closest point on κ_{1,2} to c_{0} so d(c_{0}, x) ≤ d(c_{0}, c_{1}) : l(κ_{0,4}) ≤ l(κ_{0,1}).

(iii): P_{0,1} to the right of p_{3,1} .

Again let x = P_{0,1} ∩κ_{1,2} . By inspection, ^{}c̸_{0}c_{5}x ≥ ^{}R̸(x)c_{5}x and by Claim 2, R̸(x)c_{5}x ≥ π/2 − ψ′ = ^{}c̸_{0}xc_{5} . So d(c_{0}, c_{5}) ≤ d(c_{0}, x) ≤ d(c_{0}, c_{1}) : l(κ_{0,5}) ≤ l(κ_{0,1}).

Next: c_{1} to the right of p_{3,1} .

(iv): c_{0} to the left of p_{3,1} , min{d(c_{0}, p_{3,1}), d(c_{1}, p_{3,1})} ≥ d(m_{1,2}, p_{3,1}).

Let x = m_{1,2} . By inspection, ^{}c̸_{0}c_{4}x ≥ ^{}R̸(x)c_{4}x and by Claim 2, ^{}R̸(x)c_{4}x ≥ π/2 − ψ. Again by inspection, π/2 − ψ ≥ ^{}c̸_{0}c_{4}x. So d(c_{0}, c_{4}) ≤ d(c_{0}, x). Now d(c_{0}, x) = d(c_{0}, m_{1,2}) ≤ d(c_{0}, c_{1}) : l(κ_{0,4}) ≤ l(κ_{0,1}).

(v): c_{0} to the left p_{3,1} , min{d(c_{0}, p_{3,1}), d(c_{1}, p_{3,1})} ≤ d(m_{1,2}, p_{3,1}).

Either: (a) d(c_{1}, p_{3,1}) ≤ d(c_{0}, p_{3,1}) or (b) d(c_{0}, p_{3,1}) ≤ d(c_{1}, p_{3,1}).

For (a) (respectively (b)) let x = c_{1}(R(c_{0})). By inspection, ^{}c̸_{0}c_{4}x ≥ R̸(x)c_{4}x and by Claim 2, ^{}R̸(x)c_{4}x ≥ π/2 − ψ. Again by inspection, π/2 − ψ ≥ c̸_{0}c_{4}x. So d(c_{0}, c_{4}) ≤ d(c_{0}, c_{1}) : l(κ_{0,4}) ≤ l(κ_{0,1}).

(vi): P_{0,1} between p_{3,1}, c_{1} and to the left of m_{1,2} .

Let x = P_{0,1} ∩ κ_{1,2} . If c_{3} is to the right of P_{4,3} we show that either l(κ_{3,4}) < l(κ_{0,1}) or l(η_{4}) < l(κ_{0,1}).

Suppose c_{3} is between P_{4,3} , m_{3,0} . So d(c_{4}, c_{3}) ≤ d(c_{4}, m_{3,0}). As m_{1,2} is to the right of x, d(c_{5}, m_{1,2}) ≤ d(c_{5}, x). Since c_{3} is to the right of P_{4,3} , c_{0} is to the right of P_{5,3} . We can apply the same argument as in Claim 1 since l(κ_{1,2}) ≤ l(κ_{3,4}): κ_{1,2} lies strictly to the c_{5} side of ⊥ c_{0}c_{5} . In particular d(c_{5}, x) < d(c_{0}, x). Again d(c_{0}, x) ≤ d(c_{0}, c_{1}). That is d(c_{4}, c_{3}) ≤ d(c_{4}, m_{3,0}) = d(c_{5}, m_{1,2}) ≤ d(c_{5}, x) < d(c_{0}, x) ≤ d(c_{0}, c_{1}): l(κ_{3,4}) < l(κ_{0,1}).

Next: c_{3} to the right of m_{3,0} . So ⊥ c_{3}c_{0} is to the right of P_{5,3} and κ_{1,2} lies strictly to the c_{3} side of ⊥ c_{3}c_{0} , d(c_{3}, c_{1}) < d(c_{0}, c_{1}): l(η_{4}) < l(κ_{0,1}).

So: c_{3} is stricly to the left of P_{4,3} . By inspection, ^{}c̸_{0}c_{5}x ≥ ^{}R̸(x)c_{5}x and by Claim 2, ^{}R̸(x)c_{5}x ≥ π/2 − ψ′ = ^{}c̸_{0}xc_{5} . So d(c_{0}, c_{5}) ≤ d(c_{0}, x) ≤ d(c_{0}, c_{1}): l(κ_{0,5}) ≤ l(κ_{0,1}).

(vii): P_{0,1} between p_{3,1} , c_{1} and to the right of m_{1,2} .

Let x = P_{0,1} ∩ κ_{1,2} . By inspection, ^{}c̸_{0}c_{5}x ≥ ^{}R̸(x)c_{5}x and by Claim 2, R̸(x)c_{5}x ≥ π/2 − ψ′ = ^{}c̸_{0}xc_{5} . Again it follows that l(κ_{0,5}) ≤ l(κ_{0,1}).

Figure 23. Configurations of c_{0} and c_{1}

(viii): P_{0,1} to the right of c_{1} : use the same argment as for (iii).

For each one of (i)–(viii): if l(P_{4,1}) > l(P_{5,1}) then φ < φ′′ and so l(κ_{0,4}) < l(κ_{0,1}) or l(κ_{0,4}) < l(κ_{0,1}). Likewise if l(λ_{0}) < l(λ_{2}) then d(P_{4,1}, P_{5,1}) < l(κ_{1,2}). So we may assume that l(P_{4,1}) = l(P_{5,1}) and l(λ_{0}) = l(λ_{2}), d(P_{4,1}, P_{5,1}) = l(κ_{1,2}).

Side-pairing elements of Maskit’s fundamental domain 49

Moreover l(P_{4,1}) = l(P_{5,1}) implies m_{1,2} = p_{3,1} ∩ κ_{1,2} , m_{3,0} = p_{3,1} ∩ κ_{3,0} and l(λ_{0}) = l(λ_{2}) implies ^{}c̸_{4}ℋ_{0} = π/2.

Consider (i). If l(κ_{2,5}) > l(κ_{1,2}) then φ′ + φ′′ < Ψ′ and so l(κ_{0,4}) < l(κ_{0,1}). So we may assume l(κ_{2,5}) = l(κ_{1,2}). Also if x = c_{1} is strictly to the left of p_{3,1} then c̸_{0}c_{4}x > ^{}R̸(x)c_{4}x and so l(κ_{0,4}) < l(κ_{0,1}). So we may assume c_{1} = p_{3,1} ∩ κ_{1,2} . Likewise ^{}c̸_{0}c_{4}x > ^{}R̸(x)c_{4}x if c_{0} is strictly to the left of p_{3,1} . So we may assume c_{0} = p_{3,1} ∩ κ_{3,0} .

Checking through the proof of Claim 2 we now have l(κ_{0,4}) = l(κ_{0,1}). Indeed since l(P_{4,1}) = l(P_{5,1}) the hexagon ℋ_{0} has reflective symmetries in p_{3,1} and λ_{0} . As both c_{1} = p_{3,1} ∩ κ_{1,2} , c_{0} = p_{3,1} ∩ κ_{3,0} so l(κ_{0,4}) = l(κ_{0,5}) = l(κ_{1,4}) = l(κ_{1,5}). Similarly for arcs on ℋ_{2} , which is isometric to ℋ_{0} by an orientation preserving isometry. It follows that each arc κ_{∗,∗} is the same length. As the arc set K has the combinatorial edge pattern of an octahedron, it is now not hard to show that 𝒪 must be the octahedral orbifold.

We can argue similarly for each of the other parts (ii)–(viii).

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50 David Griffiths

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Received 16 September 1996 Received in revised form 30 September 1998